•  i 
M 


SOLID   GEOMETRY 


BY 

FLETCHER  DURELL,  PH.D. 

HEAD    OF    THE    MATHEMATICAL    DEPARTMENT,    THE    LAWRENCEVII LE    SCHOOL 


NEW  YORK 
CHARLES    E.    MERRILL    CO. 


DURELL'S  MATHEMATICAL  SERIES 

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Copyright,  1904,  by  Charles  K  Merrill  Co. 


PREFACE 

ONE  of  the  main  purposes  in  writing  this  book  has 
been  to  try  to  present  the  subject  of  Geometry  so  that  the 
pupil  shall  understand  it  not  merely  as  a  series  of  correct 
deductions,  but  shall  realize  the  value  and  meaning  of  its 
principles  as  well.  This  aspect  of  the  subject  has  been 
directly  presented  in  some  places,  and  it  is  hoped  that  it  per- 
vades and  shapes  the  presentation  in  all  places. 

Again,  teachers  of  Geometry  generally  agree  that  the 
most  difficult  part  of  their  work  lies  in  developing  in 
pupils  the  power  to  work  original  exercises.  The  second 
main  purpose  of  the  book  is  to  aid  in  the  solution  of  this 
difficulty  by  arranging  original  exercises  in  groups,  each 
of  the  earlier  groups  to  be  worked  by  a  distinct  method. 
The  pupil  is  to  be  kept  working  at  each  of  these  groups 
till  he  masters  the  method  involved  in  it.  Later,  groups 
of  mixed  exercises  to  be  worked  by  various  methods  are 
given. 

In  the  current  exercises  at  the  bottom  of  the  page, 
only  such  exercises  are  used  as  can  readily  be  solved  in 
connection  with  the  daily  work.  All  difficult  originals  are 
included  in  the  groups  of  exercises  as  indicated  above. 

Similarly,  in  the  writer's  opinion,  many  of  the  nuineri- 

(iii) 

797957 


iV  PREFACE 

cal  applications  of  geometry  call  for  special  methods  of 
solution,  and  the  thorough  treatment  of  such  exercises 
should  be  taken  up  separately  and  systematically.  [See 
pp.  304-318,  etc.]  In  the  daily  extempore  work  only  such 
numerical  problems  are  included  as  are  needed  to  make 
clear  and  definite  the  meaning  and  value  of  the  geometric 
principles  considered. 

Every  attempt  has  been  made  to  create  and  cultivate 
the  heuristic  attitude  on  the  part  of  the  pupil.  This  has 
been  done  by  the  method  of  initiating  the  pupil  into 
original  work  described  above,  by  queries  in  the  course 
of  proofs,  and  also  at  the  bottom  of  different  pages,  and 
also  by  occasional  queries  in  the  course  of  the  text  where 
definitions  and  discussions  are  presented.  In  the  writer's 
opinion,  the  time  has  not  yet  come  for  the  purely  heuristic 
study  of  Geometry  in  most  schools,  but  it  is  all-important  to 
use  every  means  to  arouse  in  the  pupil  the  attitude  and 
energy  of  original  investigation  in  the  study  of  the  subject. 

In  other  respects,  the  aim  has  been  to  depart  as  little 
as  possible  from  the  methods  most  generally  used  at 
present  in  teaching  geometry. 

The  Practical  Applications  (Groups  88-91)  have  been 
drawn  from  many  sources,  but  the  author  wishes  to  ex- 
press his  especial  indebtedness  to  the  Committee  which  has 
collected  the  Real  Applied  Problems  published  from  time 
to  time  in  School  Science  and  Mathematics,  and  of  which 
Professor  J.  F.  Millis  of  the  Francis  W.  Parker  School 
of  Chicago  is  the  chairman.  Page  360  is  due  almost  en- 
tirely to  Professor  William  Betz  of  the  East  High  School 
of  Rochester,  K  Y, 

FLETCHER  DURELL. 

LAWRENCEVILLE,  N.  J.,  Sept.  1,  1904. 


TO    THE    TEACHER 

1.  IN  working  original  exercises,  one  of  the  chief  dif- 
ficulties o£  pupils  lies  in  their  inability  to  construct  the 
figure  required  and   to  make  the    particular  enunciation, 
from  it.     Many  pupils,  who  are  quite  unable  to  do  this 
preliminary  work,  after  it  is  done  can  readily  discover  a 
proof  or  a  solution.     In  many  exercises  in  this  book  the 
figure  is  drawn  and  the  particular  enunciation  made.     It 
is  left  to  the  discretion  of  the  teacher  to  determine  for 
what  other  exercises  it  is  best  to  do  this  for  pupils. 

2.  It  is  frequently  important  to  give  partial  aid  to  tho 
pupil  by  eliciting  the  outline  of  a  proof  by  questions  such 
as  the  following:   "On  this  figure  (or,  in  these  two  tri- 
angles)  what  angles  are   equal,  and  why?"    "What  lines 
are  equal,  and  why?"  etc. 

3.  In  many  cases  it  is  also  helpful  to  mark  in  colored 
crayon  pairs  of  equal  lines,  or  of  equal  angles.     Thus,  in 
the  figure  on  p.  37  lines  AB  and  DE  may  be  drawn  with 
red  crayon,  AC  and  DF  with  blue,  and  the  angles  A  and  D 
marked    by    small    arcs    drawn   with    green   crayon.      If 
colored  crayons    are  not  at  hand,  the  homologous  equal 
parts  may  be   denoted   by  like  symbols  placed   on   them, 
thus :  c  F 


or  thus:    .   .      „ 

P          B     D  P         E  A  B    D 

In  solving  theorems  concerning  proportional  lines,  it  is 
occasionally  helpful   to   denote  the   lines  in   a  proportion 


vi  TO     THE     TEACHER 

(either  given  or  to  be  proved)  by  fig- 
ures denoting  the  order  in  which  the 
lines  are  to  be  taken.  Thus,  if  OA: 
OC=OD:OB,  the  relation  may  be 
indicated  as  on  the  figure. 

4.  It  is   sometimes   helpful   to  vary 

the  symbolism  of  the  book.     Thus,   in  dealing  with  in- 
equalities a  convenient  symbol  for  "angle"  is  i£t 

as  2f.  A  >  3!  B. 

5.  Each  pupil  need  be  required  to  work  only  so  many 
originals  in  each  group  as  will  give  him  a  mastery  of  the 
particular  method  involved.     A  large  number  of  exercises 
is  given  in  order  that  the  teacher  may  have  many  to  select 
from  and  may  vary  the  work  with  successive  classes. 

6.  It  is  important  to  insist  that  the  solutions  of  exer- 
cises for  the  first  few  weeks  be  carefully  written  out;  later, 
for  many  pupils,  oral  demonstration  will  be  sufficient  and 
ground  can  be  covered  more  rapidly  by  its  use. 

7.  In    leading    pupils   to    appreciate   the    meaning   of 
theorems,  it  is  helpful  at  times  to  point  out  that  not  every 
theorem  has  for  its  object  the  demonstration  of  a  new  and 
unexpected  truth  (i.  e.,  not  all  are  "synthetic"),  but  that 
some  theorems  are  analytic,  it  being  their  purpose  to  re- 
duce an  obvious  truth  to  the  certain  few  principles  with 
which  we  start  in  Geometry.     Their  function  is,  therefore, 
to  simplify  and  clarify  the  subject  rather  than  to  extend 
its  content. 


REFERENCES   TO   PLANE   GEOMETRY. 

DEFINITIONS  AND  FIRST   PRINCIPLES. 

41.     Parallel  lines  are  lines  in  the  same  plane  which  do 
not  meet,  however  far  they  be  produced. 

GENERAL  AXIOMS. 

1.  Things  which  are  equal  to  the  same  thing,  or  to  equal 
things,  are  equal  to  each  other. 

2.  If  equals  be  added  to  equals,  the  sums  are  equal. 

3.  //  equals  be  subtracted  from  equals,  the  remainders  are 
equal. 

4.  Doubles  of  equals  are  equal;    or,  in  general,  if  equals 
be  multiplied  by  equals  the  products  are  equal. 

5.  Halves  of  equals  are  equal;  or,  in  general,  if  equals  be 
divided  by  equals  the  quotients  are  equal. 

6.  The  whole  is  equal  to  the  sum  of  its  parts. 

7.  The  whole  is  greater  than  any  of  its  parts. 

8.  A   quantity  may  be  substituted  for  its  equal  in  any 
process. 

9.  //  equals  be  added  to,  or  subtracted  from,  unequals,  the 
results  are  unequal  in  the  same  order;    if  unequals  be  added 
to  unequals  in  the  same  order,  the  results  are  unequal  in  that 
order. 

10.  Doubles,    or   halves,    of  unequals   are   unequal  in   the 
same  order. 

11.  //  unequals  be  subtracted  from  equals,  the  remainders 
are  unequal  in  the  reverse  order, 

vii 


Vlll  KEFERENCES   TO    PLANE    GEOMETRY. 

12.  //,  of  three  quantities,  the  first  is  greater  than  the.  second, 
and  the  second  is  greater  than  the  third,  then  the  first  is  greater 
than  the  third. 

GEOMETRIC   AXIOMS. 

1.  Through  two  given  points  only  one  straight  line  can  be 
passed. 

2.  A  geometric  figure  may  be  freely  moved  in  space  with- 
out any  change  in  form  or  size. 

3.  Through  a  given  point  one  straight  line  and  only  one 
can  be  drawn  parallel  to  another  given  straight  line. 

Geometric  figures  which  coincide  are  equal. 
71.     At  a  given  point  in  a  straight  line  but  one  perpen- 
dicular can  be  erected  to  the  line. 

75.  The  complements  of  two  equal  angles  are  equal;    tlu 
supplements  of  two  equal  angles  are  equal. 

76.  The  sum  of  all  the  angles  about  a  point  equals  four 
right  angles. 

BOOK  I. 

78.  //   one   straight   line  intersects   another   straight  line, 
the  opposite  or  vertical  angles  are  equal. 

79.  //,  from  a  point  in  a  perpendicular  to  a  given  line, 
two  oblique  lines  be  drawn  cutting  off  on  the  given  line  equal 
segments  from  the  foot  of  the  perpendicular,  the  oblique  lines 
are  equal  and  make  equal  angles  with  the  perpendicular. 

81.  A  triangle  is  a  portion  of  a  plane  bounded  by  three 
straight  lines. 

92.  The  sum  of  any  two  sides  of  a  triangle  is  greater  than 
the  third  side. 

94.  The   perpendicular   is   the   shortest   line   that   can   be 
drawn  from  a  given  point  to  a  given  line. 

95.  If,  from  a  point  within  a  triangle,  two  lines  be  drawn 
to  the  extremities  of  one  side  of  the  triangle,  the  sum  of  the 


REFERENCES   TO    BOOK    I.  IX 

other  two  sides  of  the  triangle  is  greater  than  the  sum  of  the 
two  lines  so  drawn. 

96.  Two  triangles  are  equal  if  two  sides  and  the  included 
angle  of  one  are  equal,  respectively,  to  two  sides  and  the  in- 
cluded angle  of  the  other. 

98.  Two  right  triangles  are  equal  if  the  hypotenuse  and  an 
acute  angle  of  one  are  equal  to  the  hypotenuse  and.  an  acute 
angle  of  the  other. 

99.  In  an  isosceles  triangle  the  angles  opposite  the  equal 
sides  are  equal. 

100.  //  two  angles  of  a  triangle  are  equal,  the  sides  oppo- 
site are  equal,  and  the  triangle  is  isosceles. 

101.  Two  triangles  are  equal  if  three  sides  of  one  are  equal 
to  three  sides  of  the  other,  respectively. 

102.  Two  right  triangles  are  equal  if  the  hypotenuse  and 
a  leg  of  one  are  equal  to  the  hypotenuse  and  a  leg  of  the  other. 

108.  //  two  sides  of  a  triangle  are  equal,  respectively,  to 
two  sides  of  another  triangle,  but  the  third  side  of  the  first  is 
greater  than  the  third  side  of  the  second,  then  the  angle  opposite 
the  third  side  of  the  first  triangle  is  greater  than  the  angle  oppo- 
site the  third  side  of  the  second. 

109.  Of  lines  drawn  from  the  same  point  in  a  perpen- 
dicular, and  cutting  off  unequal  segments  from  the  foot  of  the 
perpendicular,  the  more  remote  is  the  greater. 

112.  I.  Every  point  in  the  perpendicular  bisector  of  a  line 
is  equally  distant  from  the  extremities  of  the  line;  and 

II.  Every  point  not  in  the  perpendicular  bisector  is  un- 
equally distant  from  the  extremities  of  the  line. 

113.  Two  points  each  equidistant  from  the  extremities  of 
a  line  determine  the  perpendicular  bisector  of  the  line. 

115.  T1i-e  perpendicular  bisector  of  a  line  v'.s  tJtc  locus  of 
all  points  equidistant  from  the  extremities  of  the  line. 

120.  Parallel  lines  are  lines  in  the  same  plane  which 
do  not  meet,  however  far  they  be  produced. 


X  REFERENCES    TO    PLANE    GEOMETRY. 

121.  Two  straight  lines  in  the  same  plane,  perpendicular 
to  the  same  straight  line,  arc  parallel. 

122.  Two  straight  lines  parallel  to  a  third  straight  line 
are  parallel  to  each  other; 

Lines  parallel  to  parallel  lines  are  parallel; 

Lines  perpendicular  to  parallel  lines  are  parallel; 

Lines  perpendicular  to  non-parallel  lines  are  not  parallel. 

123.  If  a  straight  line  is  perpendicular  to  one  of  two  given 
parallel  lines  it  is  perpendicular  to  the  other  also. 

125.  //  two  straight  lines  are  cut  by  a  transversal,  mak- 
ing the  alternate  interior  angles  equal,  the  two  straight  lines 
are  parallel. 

127.  //  two  straight  lines  are  cut  by  a  transversal,  making 
the  exterior  interior  angles  equal,,  the  two  straight  lines  are 
parallel. 

130.  Two  angles  whose  sides  are  parallel,  each  to  each,  are 
either  equal  or  supplementary. 

134.  The  sum  of  the  angles  of  a  triangle  is  equal  to  two 
right  angles. 

142.  Two  right  triangles  are  equal  if  a  leg  and  an  acute 
angle  of  one  are  equal  to  a  leg  and  the  homologous  acute  angle 
of  the  other. 

147.  A  parallelogram  is  a  quadrilateral  whose  opposite 
sides  are  parallel. 

149.  A  rhombus  is  a  rhomboid  whose  sides  are  equal. 

150.  A   rectangle  is  a  parallelogram  whose  angles  are 
right  angles. 

151.  A  square  is  a  rectangle  whose  sides  are  equal. 

155.  The  opposite  sides  of  a  parallelogram  are  equal,  and 
its  opposite  angles  are  also  equal. 

156.  A  diagonal  divides  a  parallelogram  into  two  equal 
triangles. 

157.  Parallel  lines  comprehended  between  parallel  lines  are 
equal, 


REFERENCES   TO   BOOK   II.  XI 

180.  //  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  other  two  sides  are  equal  and  parallel  and  the  figure  is  a 
parallelogram. 

1 62.  Two  parallelograms  are  equal  if  two  adjacent  sides  and 
the  included  angle  of  one  are  equal,  respectively,  to  two  adjacent 
sides  and  the  included  angle  of  the  other. 

163.  Two   rectangles   which  have   equal   bases   and  equal 
altitudes  are  equal. 

174.  In  an  equiangular  polygon  of  n  sides  each  angle 

,    (n-2)2rf.Z8  2n-4 

equals ,  «  or       rt.  Z  s. 

n  n 

175.  The  sum  of  the' exterior  angles  of  a  polygon  formed 
by  producing  its  sides  in  succession  equals  four  right  angles. 

176.  //  three  or  more  parallels  intercept  equal  parts  on  one 
transversal,  they  intercept  equal  parts  on  every  transversal. 

177.  The  line  which  joins  the  midpoints  of  two  sides  of  a 
triangle  is  parallel  to  the  third  side,  and  is  equal  to  one-half 
the  third  side. 

178.  The  line  which  bisects  one  side  of  a  triangle  and  is 
parallel  to  another  side  bisects  the  third  side. 

179.  The  line  which  joins  the  midpoints  of  the  legs  of  a 
trapezoid  is  parallel  to  the  bases  and  equal  to  one-half  their 
sum. 

BOOK  II. 

197.  A  circle  is  a  portion  of  a  plane  bounded  by  a  curved 
line,  all  points  of  which  are  equally  distant  from  a  point 
within  called  the  center. 

198.  A  radius  of  a  circle  is  a  straight  line  drawn  from 
the  center  to  any  point  on  the  circumference. 

208.  Radii  of  the  same  circle,  or  of  equal  circles,  are 
equal. 

216.  In  the  same  circle,  or  in  equal  circles,  equal  arcs 
subtend  equal  angles  at  the  center. 


Xli  REFERENCES   TO    PLANE    GEOMETRY. 

217.  In  the  same  circle,  or  in  equal  circles,  of  two  unequal 
central  angles  the  greater  angle  intercepts  the  greater  arc,  and, 
conversely,  of  two  unequal  arcs,  the  greater  arc  subtends  the 
greater  angle  at  the  center. 

218.  In  the  same  circle,  or  in  equal  circles,  equal  chords  \ 
subtend  equal  arcs. 

219.  In  the  same  circle,  or  in  equal  circles,  equal  arcs  are 
subtended  by  equal  chords. 

220.  In  the  same  circle,  or  in  equal  circles,  the  greater 
of  two  (minor)  arcs  is  subtended  by  the  greater  chord;    and, 
CONVERSELY,   the  greater  of  two  chords  subtends  the  greater 
(minor)  arc. 

226.  In  the  same  circle,  or  in  equal  circles,  equal  chords 
are  equidistant  from  the  center;  and,  CONVERSELY,  chords 
which  are  equidistant  from  the  center  are  equal. 

229.  A  straight  line  perpendicular  to  a  radius  at  its  ex- 
tremity is  tangent  to  the  circle. 

230.  The  radius  drawn  to  the  point  of  contact  is  perpen- 
dicular to  a  tangent  to  a  circle. 

237.  The  two  tangents  drawn  to  a  circle  from  a  point  out- 
side the  circle  are  equal,  and  make  equal  angles  with  a  line 
drawn  from  the  point  to  the  center. 

241.  //  two  circles  intersect,  their  line  of  centers  is  per- 
pendicular to  their  common  chord  at  its  middle  point. 

253.  Properties  of  variables  and  limits. 

1.  The  limit  of  the  sum  of  a  number  of  variables  equals 
the  sum  of  the  limits  of  these  variables. 

2.  The  limit  of  a  times  a  variable  equals  a  times  the  limit 
of  the  variable,  a  being  a  constant. 

3.  The  limit  of  —th  part  of  variable  is  —th  part  of  the  limit 

1  a  a 

of  the  variable,  a  being  a  constant. 

254.  //   two   variables   are   always   equal,    and   each   ap- 
proaches a  limit,  their  limits  are  enual. 


REFERENCES   TO    BOOK    III.  Xlll 

257.  The  number  of  degrees  in  a  central  angle  equals  the 
number  of  degrees  in  the  intercepted  arc;  that  is,  a  central 
angle  is  measured  by  its  intercepted  arc. 

273.  From  a  given  point  without  a  given  line  to  draw  a 
perpendicular  to  the  line. 

274.  At  a  given  point  in  a  given  line  erect  a  perpen- 
dicular to  that  line. 

279.  Through  a  given  point  without  a  given  straight  line 
to  draw  a  line  parallel  to  a  given  line. 

286.     To  circumscribe  a  circle  about  a  given  triangle. 

BOOK    III. 

303.  The  mean  proportional  between  two  quantities  is 
equal  to  the  square  root  of  their  product. 

305.  //  the  antecedents  of  a  proportion  are  equal,  the  con- 
sequents are  equal. 

307.  //  Jour  quantities  are  in  proportion,  they  are  in  pro- 
portion by  alternation ;  that  is,  the  first  term  is  to  the  third  as 
the  second  is  to  the  fourth. 

310.  //  four  quantities  are  in  proportion,  they  arc  in  pro- 
portion  by  division ;  that  is,  the  difference  of  the  first  two  is  to 
the  second  as  the  difference  of  the  last  two  is  to  the  last. 

312.  In  a  series  of  equal  ratios,  the  sum  of  all  the  ante- 
cedents is  to  the  sum  of  all  the  consequents  as  any  one  ante- 
cedent is  to  its  consequent. 

314.  Like  powers,  or  like  roots,  of  the  terms  of  a  propor- 
tion are  in  proportion. 

317.  A  line  parallel  to  one  side  of  a  triangle  and  meeting 
the  other  two  sides,  divides  these  sides  proportionally. 

318.  If  a  line  parallel  to  the  base  cut  the  sides  of  a  triangk, 
a  side  is  to  a  segment  of  that  side  as  the  other  side  is  to  the  cor- 
responding segment  of  the  second  side. 

321.     Similar  polygons   are   polygons   having   their  ho- 


XIV  REFERENCES   TO    PLANE    GEOMETRY. 

mologous  angles  equal   and  their  homologous   sides  propor- 
tional. 

323.     //  two  triangles  are  mutually  equiangular,  they  are 
similar. 

326.  //  two  triangles  have  their  homologous  sides  propor- 
tional they  are  similar. 

327.  //  two  triangles  have  an  angle  of  one  equal  to  an  angle 
of  the  other,  and  the  including  sides  proportional,  the  triangles 
are  similar. 

328.  //  two  triangles  have  their  sides  parallel,  or  perpen- 
dicular, each  to  each,  the  triangles  are  similar. 

329.  //  two  polygons  are  similar,  they  may  be  separated 
into  the  same  number  of  triangles,  similar,  each  to  each,  and 
similarly  placed. 

342.  In  a  right  triangle, 

I.  The  altitude  to  fhe  hypotenuse  is  a  mean  proportional 
between  the  segments  of  the  hypotenuse; 

II.  Each   leg  is   a  mean   proportional  between   the  hypote- 
nuse and  the  segment  of  the  hypotenuse  adjacent  to  the  given 
leg. 

343.  The  perpendicular  to  the  diameter  from  any  point 
in  the  circumference  of  a  circle  is  a  mean  proportional  between 
the  segments  of  the  diameter;    and  the  chord  joining  the  point 
to  an  extremity  of  the  diameter  is  a  mean  proportional  be- 
tween the  diameter  and  the  segment  of  the  diameter  adjacent  to 
the  chord. 

346.  In  a  right  triangle,  the  square  of  the  hypotenuse  is 
equal  to  the  sum  of  the  squares  of  the  legs. 

347.  In  a  right  triangle,  the  square  of  either  leg  is  equal 
to  the  square  of  the  hypotenuse  minus  the  square  of  the  other 
leg. 

351.  //  the  square  on  the  side  of  a  triangle  equals  the  sum 
of  the  squares  on  the  other  two  sides,  the  angle,  opposite  the  first 
side  is  a  right  angle. 


REFERENCES   TO   BOOK   IV.  XV 

352.     //,  in  any  triangle,  a  median  be  drawn  to  one  side, 

I.  The  sum  of  the  squares  of  the  other  two  sides  is  equal 
to  twice  the  square  of  half  the  given  side,  increased  by  twice 
the  square  of  the  median  upon  that  side;  and 

II.  The  difference  of  the  squares  of  the  other  two  sides  is 
equal  to  twice  the  product  of  the  given  side  by  the  projection 
of  the  median  upon  that  side. 

BOOK   IV. 

383.  The  area  of  a  rectangle  is  equal  to  the  product  of  its 
base  by  its  altitude. 

385.  The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  base  by  its  altitude. 

389.  The  area  of  a  triangle  is  equal  to  one-half  the  prod- 
uct of  its  base  by  its  altitude. 

390.  Triangles   which   have   equal   bases   and   equal  alti- 
tudes (or  which  have  equal  bases  and  their  vertices  in  a  line 
parallel  to  the  base)  are  equivalent. 

391.  Triangles  which  have  equal  bases  are  to  each  other  as 
their  altitudes; 

Triangles  which  have  equal  altitudes  are  to  each  other  as 
their  bases. 

392.  Any  two  triangles  are  to  each  other  as  the  products 
of  their  bases  and  altitudes. 

397.  //  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other,  their  areas  are  to  each  other  as  the  products 
of  the  sides  including  the  equal  angles. 

398.  The  areas  of  any  two  similar  triangles  are  to  each 
other  as  the  squares  of  any  two  homologous  sides. 

399.  The  areas  of  two  similar  polygons  are  to  each  other 
as  the  squares  of  any  two  homologous  sides. 


XVI  REFERENCES   TO    PLANE   GEOMETRY. 

BOOK    V. 

444.     Formula  for  the  circumference  in  terms  of  the 
radius. 


449.     The  area  of  a  circle  is  equal  to  one-half  the  product 
of  its  circumference  by  its  radius. 

486.  Two  points  symmetrical  with  respect  to  a  line  or 
axis  are  points  such  that  the  straight  line  joining  them  is 
bisected  by  the  given  line  at  right  angles. 

487.  A  figure  symmetrical  with  respect  to  an  axis  is  a 
figure  such  that  each  point  in  the  one  part  of  the  figure  has 
a  point  in  the  other  part  symmetrical  to  the  given  point, 
with  respect  to  an  axis. 

489.  Two  points  symmetrical  with  respect  to  a  point  or 
center  are  points  such  that  the  straight  line  joining  them  is 
bisected  by  the  point  or  center. 

490.  A  figure  symmetrical  with  respect  to  a  point  or 
center  is  a  figure  such  that  each  point  in  the  figure  has  an- 
other point  in  the  figure  symmetrical  to  the  given  point  with 
respect  to  the  center. 


TABLE  OF  CONTENTS 

PAGE 

BOOK  VI.    LINES,  PLANES  AND  ANGLES  IN  SPACE 319 

BOOK  VII.    POLYHEDRONS 360 

BOOK  VIII.    CYLINDERS  AND  CONES .   .  .  403 

BOOK  IX.     THE  SPHERE 425 

NUMERICAL  EXERCISES  IN  SOLID  GEOMETRY 4(59 

APPENDIX  I.     MODERN  GEOMETRIC  CONCEPTS 485 

II.     HISTORY  OF  GEOMETRY 490 

III.    REVIEW  EXERCISES.     PLANE  GEOMETRY    ....  491 

SOLID  GEOMETRY 506 

PRACTICAL  APPLICATIONS  OF  SOLID  GEOMETRY 510 

FORMULAS  .  516 


(rvii) 


SYMBOLS    AND    ABBREVIATIONS 

H-  plus,  or  increased  by.  Adj.    .   .  adjacent. 

—  minus,  or  diminished  by.  Alt.     .   .  alternate. 
X  multiplied  by.  Art.     .   .  article. 
-s-  divided  by.  Ax.  ...  axiom. 

=  equals;  is  (or  are]  equal  to.  Constr.  .  construction0 

—  approaches  (as  a  limit).  Cor.     .   .  corollary. 
=c=  is  (or  are)  equivalent  to.  Def.     .   .  definition. 
>  is  (or  are)  greater  than.  Ex.  .   .    .  exercise. 
<  is  (or  arc)  Zess  Ma?i.  Ext.     .   .  exterior. 
.'.  therefore.  Fig.     .   .  figure. 

J_  perpendicular,  perpendicular  to,      Hyp.    .  .  hypothesis. 

or,  is  perpendicular  to.  Ident.  .  identity. 

.k  perpendiculars.  Int.      .  .  interior. 

||    parallel,  or,  is  parallel  to.  Post.   .  .  postulate. 

\\sparallels.  Prop.  .  .proposition. 

/. ,  A  angle,  angles  Rt.  .   .  .  right. 

A,  A  triangle,  triangles.  Sug.     .  .  suggestion. 

/    7  ,  £Z7  parallelogram,  parallelograms.         Sup.    .  .  supplementary. 

G,  ©  cirete,  circles.  St.    ...  straight. 

Q.  E.  D.  quod    erat    demonstrandum;    that    is,    which    was   to    be 

proved. 
Q.  E.  F.  quod  erat  faciendum;  that  is,  which  was  to  be  made. 

A  few  other  abbreviations  and   symbols  will   be  introduced   and 
their  meaning  indicated  later  on. 


SOLID  GEOMETRY 

BOOK  VI 
LINES,    PLANES   AND   ANGLES   IN   SPACE 

DEFINITIONS    AND   FIRST   PRINCIPLES 

497.  Solid  Geometry  treats  of  the  properties  of  space  of 
three  dimensions. 

Many  of  the  properties  of  space  of  three  dimensions  are  determined 
by  use  of  the  plane  and  of  the  properties  of  plane  figures  already 
obtained  in  Plane  Geometry. 

498.  A  plane  is  a  surface  such  that,  if  any  two  points 
in  it  be  joined  by  a  straight  line,  the  line  lies  wholly  in  the 
surface. 

499.  A  plane  is  determined  by  given  points  or  lines,  if 
no  other  plane  can  pass  through  the  given  points  or  lines 
without  coinciding  with  the  given  plane. 

500.  Fundamental  property  of  a  plane  in  space.      A 
plane  is  determined  by  any  three  points  not  in  a  straight  line. 

For,  if  through  a  line  con-  .c 

necting  two  given  points,  A 
and  B,  a  plane  be  passed,  the 
plane,  if  rotated,  can  pass 
through  a  third  given  point, 
(7,  in  but  one  position. 

The  importance  of  the  above  principle  is  seen  from  the  fact  that  it 
reduces  an  unlimited  surface  to  three  points,  thus  making  a  vast 
economy  to  the  attention.  It  also  enables  us  to  connect  different 
planes,  and  treat  of  their  properties  systematically. 

(319) 


320  BOOK   VI.       SOLID     GEOMETRY 

501.  Other  modes  of  determining  a  plane.  A  plane  may 
also  be  determined  by  any  equivalent  of  three  points  not  in 
a  straight  line,  as  by 

a  straight  line  and  a  point  outside  the  line;  or  by 
two  intersecting  straight  lines;  or  by 
two  parallel  straight  lines. 

It  is  often  more  convenient  to  use  one  of  these  latter  methods  of 
determining  a  plane  than  to  reduce  the  data  to  three  points  and  use 
Art.  500. 


\ 


502.  Representation  of  a  plane  in  geometric  figures.    la 
masoning  concerning  the  plane,  it  is  often  an  advantage 
to  have  the  plane  represented  in  all  directions.     Hence,  in 
drawing  a  geometric  figure,  a  plane  is  usually  represented 
to  the  eye  by  a  small  parallelogram. 

This  is  virtually  a  double  use  of  two  intersecting  lines,  or  of  two 
parallel  lines,  to  determine  a  plane  (Art.  501). 

503.  Postulate  of  Solid  Geometry.      The  principle  of 
Art.  499  may  also  be  stated  as  a  postulate,  .thus: 

Throtigh  any  three  points  not  in  a  straight  line  (or  their 
equivalent)  a  plane  may  be  passed. 

504.  The  foot  of  a  line  is  the  point  in  which  the  line 
intersects  a  given  plane. 

505.  A  straight  line  perpendicular  to  a  plane  is  a  line 
perpendicular  to  every  line  in  the  plane  drawn  through  its 
foot. 

A  straight  line  perpendicular  to  a  plane  is  sometimes  called  a  normal 
to  the  plane. 


LINES    AND    PLANES  321 

506.  A  parallel  straight  line  and  plane  are  aline  and 
plane  which  cannot  meet,  however  far  they  be  produced. 

507.  Rarallel  planes  are   planes   which   cannot   meet, 
however  far  they  be  produced. 

508.  Properties  of  planes  inferred  immediately. 

1.  A  straight  line,  not  in  a  given  plane,  can  intersect  the 
given  plane  in  but  one  point. 

For,  if  the  line  intersect  the  given  plane  in  two  or  more 
points,  by  definition  of  a  plane,  the  line  must  lie  in  the 
plane.  Art.  498. 

2.  The  intersection  of  two  planes  is  a  straight  line. 
For,  if  two  points  common  to  the  two  planes  be  joined 

by  a  straight  line,  this  line  lies  in  each  plane  (Art.  498) ; 
and  no  other  point  can  be  common  to  the  two  planes,  for, 
through  a  straight  line  and  a  point  outside  of  4t  only  one 
plane  can  be  passed.  Art.  501. 


Ex.  1.  Give  an  example  of  a  plane  surface;  of  a  curved  surface; 
of  a  surface,  part  plane  and  part  curved;  of  a  surface  composed  of 
different  plane  surfaces. 

Ex.  2.  Four  points,  not  all  in  the  same  plane  determine  how 
many  different  planes  ?  how  many  different  straight  lines  ? 

Ex.  3.  Three  parallel  straight  lines,  not  in  the  same  plane,  deter- 
mine how  many  different  planes  ? 

Ex.  4.  Four  parallel  straight  lines  can  determine  how  many  differ- 
ent planes  ? 

Ex.  5.  Two  intersecting  straight  lines  and  a  point,  not  in  their 
plane,  determine  how  many  different  planes  f 


322  BOOK   VI.      SOLID     GEOMETRY 

• 

PROPOSITION  I.     THEOREM 

509.  If  a  straight  line  is  perpendicular  to  each  of  two 
other  straight  lines  at  their  point  of  intersection,  it  is  per- 
pendicular to  the  plane  of  those  lines. 


Given  AB  _L  lines  EC  and  ED,  and  the  plane  MN  pass- 
ing through  EG  and  ED. 

To  prove  AB  _L  plane  MN. 

Proof.    Through  E  draw  EG,  any  other  line  in  the  plane 
MN. 

Draw  any  convenient  line  CD  intersecting  EC,  EG  and 
ED  in  the  points  C,  G  and  D,  respectively. 
Produce  the  line  AB  to  F,  making  BF=AB. 

Connect  the  points  C,  G,  D  with  A,  and  also  with  F. 

Then,  in  the  A  ACD  and  FCD,  CD=CD.  Ident. 

A  C=  CF,  and  AD=DF.  Art.  112. 

/.  &ACD=AFCD.  (Why?) 

/.    Z  ACD=Z.  FCD.  (Why?) 

Then,  in  the  A  ACG  and  FCG,  CG=CG,  (Why?) 

AC=  CF,  and  Z  ACG  =  Z JW#.  (Why ?) 

.'.   A  J.C#=A  ^(7O.  (Why?) 


LINES     AND    PLANES  323 

/.  AG=GF.  (Why?) 

.*.  B  and  O  are  each  equidistant  from  the  points  A  and  F. 
:.  BO  is  J_  AF;  that  is,  AB  J_  BO.         Art.  113. 

/.  AB  _L  plane  MN,  Art.  505. 

(for  it  is  J_  any  line,  EG,  in  the  plane  MN,  through  its  foot}. 

Q.  E.  D. 

PROPOSITION  II.     THEOREM 

510.  All  the  perpendiculars  that  can  be  drawn  to  a 
given  line  at  a  given  point  in  the  line  lie  in  a  plane  perpen- 
dicular to  the  line  at  the  given  point. 


Given  the  plane  MN  and  the  line  #<7both  J_  line  AB  at 
the  point  B. 

To  prove  that  BC  lies  in  the  plane  MN. 
Proof.    Pass  a  plane  AF  through  the  intersecting  lines 
AB  and  BC.  Art.  503. 

This  plane  will  intersect  the  plane  MN  in  a  straight 
line  BF.  Art.  508,  2. 

But  AB  J_  plane  MN  (Hyp.)   .-.  AB  J_  BF.          Art.  505. 
Also  AB  _L  BC.  Hyp. 

.*.  in  the  plane  AF,  BC  and  BF  _L  AB  at  B. 

:.  BC  and  BF  coincide.  Art.  71. 

But  BF  is  in  the  plane  MN. 

:.  BC  must  be  in  the  plane  MN, 
(for  BC  coincides  with  BF)  which  lies  in  the  plane  MN}. 

Q.  £•  D* 


324  BOOK  VI.       SOLID     GEOMETKY 

511.  COR.  1.  At  a  given  point  B  in  the  straight  line  AB, 
to  construct  a  plane  perpendicular  to  the  line  AB.  Pass  a 
plane  AF  through  AB  in  any  convenient  direction,  and  in 
the  plane  AF  at  the  point  B  construct  BF  J_  AB  (Art. 
274).  Pass  another  plane  through  AB,  and  in  it  construct 
BP  ±  AB.  Through  the  lines 
BF  and  BP  pass  the  plane 
MN  (Art.  503).  MN  is  the 
required  plane  (Art.  509). 


512.  COR.  2.    Through   a 
given  external  point,  P,  to  pass 
a   plane    perpendicular    to    a 

given  line,  AB.  Pass  a  plane  through  AB  and  P  (Art. 
503),  and  in  this  plane  draw  PB  _L  AB  (Art.  273).  Pass 
another  plane  through  AB,  as  AF,  and  in  AF  draw  BF  JL 
AB  at  B  (Art.  274).  Pass  a  plane  through  BP  and  BF 
(Art.  503).  This  will  be  the  plane  required  (Art.  509). 

513.  COR.  3.   Through  a  given  point  but  one  plane  can 
be  passed  perpendicular  to  a  given  line. 


Ex.  1.    Five  points,  no  four  of  which  are  in  the  same  plane,  deter- 
mine how  many  different  planes  ?  how  many  different  straight  lines  ? 

Ex.  2.    A  straight  line  and  two  points,  not  all  of  which  are  in  the 
same  plane,  determine  how  many  different  planes  ? 

Ex.  3.    In  the  figure,  p.  322,  prove  that  the  triangles  GAD  and 
GDF  are  equal. 

Ex.  4.    In  the  same  figure,  if  AB=S  and  BC=Q,  find  FC. 


LINES    AND     PLANES  325 

PROPOSITION  III.    PROBLEM    • 

514.    At  a  given  point  in  a  plane,  to  erect  a  perpendicu- 
lar to  the  plane. 

p — 


Given  the  point  A  in  plane  MN. 

To  construct  a  line  perpendicular  to  MN  at  the  point  A. 

Construction.  Through  the  point  A  draw  any  line  CD  in 
the  plane  MN. 

Also  through  the  point  A  pass  the  plane  PQ  JL  CD 
(Art.  511),  intersecting  the  plane  MN  in  the  line  RS.  Art.  508,  2. 

In  the  plane  PQ  draw  AK  J_  line  RS  at  A.         Art.  274. 

Then  AK  is  the  J_  required. 

Proof.  CD  _L  plane  PQ.  Constr. 

/.  CD  J_  AK.  Art.  505. 

Hence  AK  J_  CD. 

But  AK  _L  #£.  Constr. 


/.  AK  JL  plane  MF.  Art.  509. 

Q.  E.  F. 

515.  COR.  At  a  given  point  in  a  plane  but  one  perpen- 
dicular to  the  plane  can  be  drawn.  For,  if  two  Ja.  could  be 
drawn  at  the  given  point,  a  plane  could  be  passed  through 
them  intersecting  the  given  plane.  Then  the  two  Ja  would 
be  in  the  new  plane  and  J_  to  the  same  line  (the  line  of 
intersection  of  the  two  planes,  Art,  505),  which  is  im- 
possible (Art,  71), 


326  BOOK  VI.      SOLID    GEOMETBY 

PROPOSITION  IV.     PROBLEM 

516.    From  a  given  point  without  a  plane,  to  draw  a  line 
perpendicular  to  the  plane. 

A 

^ 


,B 


Given  the  plane  MN  and  the  point  A  external  to  it. 

To  construct  from  A  a  line  JL  plane  MN. 

Construction.  In  the  plane  MN  draw  any  convenient 
line  EC.  Pass  a  plane  through  EG  and  A  (Art.  503),  and 
in  this  plane  draw  AD  _L  EC.  Art.  273. 

In  the  plane  MN  draw  LD  J_  EC.  Art.  274. 

Pass  a  plane  through  AD  and  LD  (Art.  503),  and  in 
that  plane  draw  AL  J_  LD.  Art.  273. 

Then  AL  is  the  _L  required. 

Proof.  Take  any  point  C  in  EC  except  D,  and  draw  LC 
and  AC. 

Then  A  J.DC,  AD£  and  LDC  are  right  A.  Constr. 

.  Art.  400. 

+  ~D&.  Art.  400,  Ax.  8. 
.'.  AC2  =  AL2  +  LC2.  Art.  400,  Ax.  8. 

/.   /.ALC  is  a  right  Z  .  Art.  351. 

But  AL  JL  LD.  Constr. 

/.  AL  X  JOT.  Art.  509. 

Q.  E.  F. 

517r  COR.  But  one  perpendicular  can  be  drawn  from  a 
given  external  point  to  a  given  plane. 


LINES     AND     PLANES 


327 


PROPOSITION  V.     THEOREM     „ 

518.  I.  Oblique  lines  drawn  from  a  point  to  a  plane, 
meeting  the  plane  at  equal  distances  from  the  foot  of  the 
perpendicular,  are  equal; 

II.  Of  two  oblique  lines  drawn  from  a  point  to  a  plane, 
but  meeting  the  plane  at  unequal  distances  from  the  foot  of 
the  perpendicular,  the  more  remote  is  the  greater. 


Given  AB  i.  plane  MN,  BD  =  BC,  and  BH  >  EG. 
To  prove     AD  =  AC,  and  AH  >  AC. 
Proof.    I.    In  the  right  A  ABD  and  ABC, 
AB  =  AB,  and  BD=BC. 
/.  AABD=AABC. 


(Why?) 
(Why?) 
(Why?) 


II.  On  .RET  take  BF=BG,  and  draw  AF. 
Then  A  F=  AC  (by  part  of  theorem  just  proved) 
But  AH  >  AF. 

:.  AH  >  AC. 


(Why  ?) 

Ax.  8. 
Q.  E.  D. 


519.  COR.  1.  CONVERSELY:  Equal  oblique  lines  drawn 
from  a  point  to  a  plane  meet  the  plane  at  equal  distances 
from  the  foot  of  the  perpendicular  drawn  from  the  same  point 
to  the  plane;  and,  of  two  unequal  lines  so  drawn,  the  greater 
line  meets  the  plane  at  the  greater  distance  from  the  foot  of 
the  perpendicular. 


328  BOOK  VI.      SOLID    GEOMETRY 

520.  COR.  2.    The  locus  of  a  point  in  space  equidistant 
from  all  the  points  in   the  circumference  of  a  circle  is  a 
straight  line  passing  through  the  center  of  the  circle  and 
perpendicular  to  its  plane. 

521.  COB.  3.    The  perpendicular  is  the  shortest  line  that 
can  l)e  drawn  from  a  given  point  to  a  given  plane. 

522.  DEF.    The  distance  from  a  point  to  a  plane  is  the 
perpendicular  drawn  from  the  point  to  the  plane. 

PROPOSITION  VI.     THEOREM 

523.  If  from  the  foot  of  a  perpendicular  to  a  plane  a 
line  l)e  drawn  at  right  angles  to  any  line  in  the  plane,  the  line 
drawn  from  the  point  of  intersection  so  formed  to  any  point  in 
the  perpendicular,  is  perpendicular  to  the  line  of  the  plane. 


\A-7P 


Given  AB  J_  plane  MN,  and  EF  _L  CD,  any  line  in  MN. 
To  prove  AF  _L  CD. 

Proof.    On  CD  take  FP  and  FQ  equal  segments. 
Draw  AP,  BP,  AQ,  BQ. 

Then  BP=BQ.  Art.  112. 

Hence  AP=AQ.  Art.  518. 

/.  in  the  line  AF,  the  point  A  is  equidistant  from  P 
and  Q,  and  F  is  equidistant  from  P  and  Q. 

:.  AF  i.  CD.  (Why?) 

Q.  E.  D. 

Ex.    In  the  above  figure,  if  AS=Q,  AF=8,  and  ^£=10,  find  QF, 
BF  and  £. 


LINES    AND    PLANES  329 

PROPOSITION  VII.     THEOREM 

524.    Two  straight  lines  perpendicular  to  the  same  plane 
are  parallel. 

A  C 

\ 


•F 
D 


Given  the  lines  AB  and  CD  J.  plane  MN. 
To  prove  AB  \\  CD. 

Proof.    Draw  BD,  and  through  D,  in  the  plane 
draw  FH  J_  BD. 
Draw  AD. 
Then  BD  _L  FIT.  Constr. 

AD  J_  FH.  Art.  523. 

CD  J_  J^ff.  Art.  505. 

/.  BD,  AD  and  CD  are  all  J.  FH  at  the  point  D. 

.'.  BD,  AD  and  CD  all  lie  in  the  same  plane.       Art.  510. 

.'.  AB  and  CD  are  in  the  same  plan^.          (Why?) 

But  AB  and  CD  are  J_  #Z).  Art.  505. 

/.  AB  and  CD  are  ||.  Art.  121. 

Q.  £.  D. 

525.    COR.  1.    If  one  of  two  parallel  lines  is  perpendicu- 
lar to  a  plane,  the  other  is  perpendicular  to  the  plane  also. 

For,  if  AB  and  CD  be  ||,  and  AB  _L 
plane  PQ,  a  line  drawn  from  C  J_  PQ 
must  be  ||  AB.  Art.  524. 

But  CD  must  coincide  with  the  line 
so  drawn  (Art.  47,  3);  .-.  CD  X  PQ. 


330  BOOK  VI.       SOLID     GEOMETKY 

526.  COE.  2.  If  two  straight  lines  are  each  parallel  to 
a  third  straight  line,  they  are  parallel  to  each  other.  For, 
if  a  plane  be  drawn  _L  to  the  third  line,  each  of  the  two 
other  lines  must  be  _L  to  it  (Art.  525),  and  therefore  be  || 
to  each  other  (Art.  524). 


PROPOSITION  VIII.     THEOREM 

527.  If  a  straight  line  external  to  a  given  plane  is  paral- 
lel to  a  line  in  the  plane,  then  the  first  line  is  parallel  to  the 
given  plane. 


Given  the  straight  line  AB  \\  line  CD  in  the  plane  MN. 

To  prove  AB  \\  plane  MN. 

Proof.    Pass  a  plane  through  the  ||  lines  AB  and  CD. 

If  AB  meets  MN  it  must  meet  it  in  the  line  CD. 

But  AB  and  CD  cannot  meet,  for  they  are  ||.       Art.  120. 

.'.  AB  and  MN  cannot  meet  and  are  parallel.       Art.  506. 

Q.  E.  D. 

528.  COR.  1.  If  a  straight  line  is  parallel  to  a  plane , 
the  intersection  of  the  plane  with  any  plane  passing  through 
the  given  line  is  parallel  to  the  given  line. 


LINES     AND    PLANES 


331 


529.  COR.  2.  Through  a  given  line 
(CD)  to  pass  a  plane  parallel  to  another 
given  line  (AB) . 

Through  P,  any  point  in  CD,  draw 
QR  ||  AB  (Art.  279) .     Through  CD  and 
QR  pass  a   plane    (Art.    503).     This   will   be   the    plane 
required  (Art.  527). 

If  AB  and  CD  are  not  parallel,  but  one  plane  can  be 
drawn  through  CD\\AB. 


PROPOSITION  IX.     THEOREM 

530.    Two  planes  perpendicular  to  the  same  straight  line 
are  parallel. 


M 


Given  the  planes  MN  and  PQ  _L  line  AB. 
To  prove  MN\\  PQ. 

Proof.    If  MN  and  PQ  are  not  parallel,   on  being  pro- 
duced they  will  meet. 

We  shall  then  have  two  planes  drawn  from  a  point  per- 
pendicular to  a  given  line,  which  is  impossible.        Art.  513. 


.*.  MN  and  PQ  are  parallel. 


Art.  507. 
Q.  £.  D. 


332  BOOK  VI.      SOLID    GEOMETRY 

•          PROPOSITION  X.    THEOREM 

531.    //  two  parallel  planes  are  cut  by  a  third  plane,  the 
intersections  are  parallel  lines. 


Given  MN  and  PQ  two  1 1  planes  intersected  by  the  plan* 
RS  in  the  lines  AB  and  CD. 

To  prove  AB  \\  CD. 

Proof.   AB  and  CD  lie  in  the  same  plane  R8. 

Also  AB  and  CD  cannot  meet;  for  if  they  did  meet 
the  planes  MN  and  PQ  would  meet,  which  is  impossible. 

Art.  507. 

.*.  AB  and  CD  are  parallel.  Art.  41. 

Q.  E.  D. 

532.  COR.  1.  Parallel  lines  included  between  parallel 
planes  are  equal.  For,  if  AC  and  BD  are  two  parallel  lines, 
a  plane  may  be  passed  through  them  (Art.  503),  intersect- 
ing MN  and  PQ  in  the  ||  lines  AB  and  CD.  Art.  531. 

/.  ABDC  is  a  parallelogram.  Art.  147. 

.*.  AC=BD.  Art.  155. 


LINES     AND     PLANES 


333 


533.  COR.  2.  Two  parallel  planes  are  everywhere  equi- 
distant. 

For  lines  _L  to  one  of  them  are  ||  (Art.  524).  Hence 
the  segments  of  these  lines  included  between  the  ||  planes 
are  equal  (Art.  532). 


PROPOSITION  XI.  *  THEOREM 

534.    If  two  intersecting  lines  are  each  parallel  to  a  given 
plane,  the  plane  of  these  lines  is  parallel  to  the  given  plane. 


D 


K 


Given  the  lines  AB  and  CD,  each  ||  plane  PQ,  and  inter- 
secting in  the  point  F\  and  MN  a  plane  through  AB  and 
CD. 

To  prove  MN\\PQ. 

Proof.    From  the  point  F  draw  FH  _L  PQ. 

Pass  a  plane  through  FC  and  FH,  intersecting  PQ  in 
HK;  also  pass  a  plane  through  FB  and  Fit,  intersecting 
PQ  in  HL. 

Then  HK  \\  FC,  and  HL  \\  FB.  Art.  528. 

But  FH  _L  HK  and  HL.  Art.  505. 

/.  FH  J_  FC  and  FB.  Art.  123. 

/.  FH  _L  M^.  Art.  509. 

.-.  MN  II  P#.  Art.  530. 

Q.  B.  ». 


334 


» 


BOOK  VI.       SOLID     GEOMETRY 


PROPOSITION  XII.     THEOREM 


535.    A  straight  line  perpendicular  to  one  of  two  parallel 
planes  is  perpendicular  to  the  other  also. 


\    T" 


Given  the  plane  MN  \\  plane  PQ,  and  AB  JL  PQ. 
To  prove  AB  J_  MN. 

Proof.  Through  AB  pass  a  plane  intersecting  PQ  and 
MN  in  the  lines  BO  and  AF,  respectively;  also  through 
AB  pass  another  plane  intersecting  PQ  and  MN  in  BJ)  and 
AH,  respectively. 

Then  BO  1 1  AF,  and  J5D  ||  AJ3".  Art.  531. 

But  AB  _L  BG  and  J9D.  Art.  505. 

.*.  AB  J_  AJP  and  AH.  Art.  123. 

.*.  A-B  _L  plane  M2V.  Art.  509. 

Q.  £.  D. 

536.  COR.  1.    Through  a   given  point   to  pass  a  plane 
parallel  to  a  given  plane. 

Let  the  pupil  supply  the  construction. 

537.  COR.  2.    Through  a  given  point  but  one  plane  can 
be  passed  parallel  to  a  given  plane, 


LINES     AND     PLANES  335 

PROPOSITION  XIII.     THEOREM 

538.  If  two  angles  not  in  the  same  plane  have  their 
corresponding  sides  parallel  and  extending  in  the  same  direc- 
tion, the  angles  are  equal  and  their  planes  are  parallel. 


\ 


\ 


Given  the  /.BAG  in  the  plane  MN,  and  the  £B'A'C> 
in  the  plane  PQ;  AB  and  A'B'  \\  and  extending  in  the 
same  direction;  and  AC  and  AC'  \\  and  extending  in  the 
same  direction. 

To  prove  £BAC  =  /.B'A'C',  and  plane  MN  \\  plane  PQ. 
Proof.    Take  AB  =  ABr,  and  AC=ACf. 
Draw  AA,  BB't  CC',  BC,  B'C'. 

Then  ABB' A  is  a  ZZ7 ,  Art.  160. 

(for  AB  and  AfB'  are=and  \\  ). 

/.  BB'  and  A  A'  are  =  and  IK  Art.  155. 
In  like  manner  CC'  and  AA'  are  =  and  ||. 

/.  BB'  and  CC'  are  =  and  ||.  (Why?) 

.'.  BCC'B'  is  a  C3  ,  and  BC=B'Cf.  (Why  ?) 

/.  A  ABC=  A  A'B'C'.  (Why  ?) 

/.   £A  =  /.A.  (Why?) 

Also           AB  ||  A'B',  :.  AB  \\  plane  PQ.  Art.  527. 

Similarly  AC  \\  plane  P^. 

:.  plane  JOT  ||  plane  PQ.  Art.  534. 

o.  E.  B. 


336  BOOK  VI.       SOLID     GEOMETRY 

A     PROPOSITION  XIV.     THEOREM 

539.  //  two  straight  lines  are  intersected  by  three  paral- 
lel planes^  the  corresponding  segments  of  these  lines  are 
proportional. 

Mr 


wrp. 


Given  the  straight  lines  AB  and  CD  intersected  by  the 
||  planes  MN,  P^/and  ES  in  the  points  A,  F,  B,  and  C 
H,  D,  respectively. 

AF    GH 
To  prove 


Proof.    Draw  the  line  AD  intersecting  the  plane  PQ  in  G. 

Draw  FG,  BD,  GH,  AC. 

Then  FG  \\  BD,  and  GH  \\  AC.  Art.  531. 


Art   317 


And 

AF    CH 


(Why?) 

"  FB    HD 

Q.  E.  D. 


Ex.  1.    In  above  figure,  if  AF=2,  FB=5,  and  CH=3,  find  CD. 
Ex  2.    If  CH=3,    HD=±,  and  AB=IQ,  find  AF  and  BF. 


DIHEDKAL    ANGLES  337 

DIHEDRAL   ANGLES 

540.  A  dihedral  angle  is  the  opening  between  two  in- 
tersecting planes. 

From  certain  points  of  view,  a  dihedral  angle  may  be  regarded  as 
a  wedge  or  slice  of  space  cut  out  by  the  planes  forming  the  dihedral 
angle. 

541.  The  faces  of  a  dihedral  angle  are  the  planes  form- 
ing the  dihedral  angle. 

The  edge  of  a  dihedral  angle  is  the  straight  line  in  which 
the  faces  intersect. 

542.  Naming  dihedral  an- 
gles.    A  dihedral  angle  may  be 
named,  or  denoted,  by  naming 
its  edge,  as  the  dihedral  angle 
AB;  or  by  naming  four  points, 
two  on  the  edge  and  one  on 
each  face,   those  on  the  edge 
coming  between  the  points  on 
the  faces,  as  P-AB-Q.      The 

latter  method  is  necessary  in  naming  two  or  more  dihedral 
angles  which  have  a  common  edge. 

543.  Equal  dihedral  angles  are  dihedral  angles  which 
can  be  made  to  coincide. 

544.  Adjacent  dihedral  angles  are  dihedral  angles  hav- 
ing the  same  edge  and  a  face  between  them  in  common. 

545.  Vertical  dihedral  angles  are  two  dihedral  angles 
having  the  same  edge,  and  the  faces  of  one  the  prolonga- 
tions of  the  faces  in  the  other. 

546.  A  right  dihedral  angle  is  one  of  two  equal  adja- 
cent dihedral  angles  formed  by  two  planes. 


338 


BOOK   VI.       SOLID     GEOMETRY 


547.  A   plane   per- 
pendicular  to   a   given 
plane  is  a  plane  form- 
ing   a    right    dihedral 
angle   with    the   given 
plane. 

Many  of  the  properties  of  dihedral  angles  are  obtained 
most  conveniently  by  using  a  plane  angle  to  represent  the 
dihedral  angle. 

548.  The  plane  angle  of  a  dihedral  an- 
gle is  the  angle  formed  by  two  lines  drawn 
one  in  each  face,  perpendicular  to  the  edge 
at  the  same  point. 

Thus,  in  the  dihedral  angle  C-AB-F, 
if  PQ  is  a  line  in  the  face  AD  perpendicu- 
lar to  the  edge  AB  at  P,  and  PR  is  a  line 
in  face  AF  perpendicular  to  the  edge  AB 
at  P,  the  angle  QPR  is  the  plane  angle  of  the  dihedral 
angle  C-AB-F. 

549.  Property  of  plane  angles  of  a  dihedral  angle. 

The  magnitude  of  the  plane  angle  of  a  dihedral  angle  is 
the  same  at  every  point  of  the  edge.    For  let  EAC  be  the 
plane  Z  of  the  dihedral  Z  E-AB-D  at  the  point  A. 
Then  PR  \\  AE,  and  PQ  \\  AC  (Art.  121.) 
=  /.EAC  (Art.  538). 


D 


550.  The  projection  of  a  point  upon  a  plane  is  the  foot  of 
a  perpendicular  drawn  from  the  point  to  the  plane. 

ritt 

551.  The    projection   of    a    line 
upon  a  plane  is  the  locus  of  the  pro- 
jections of  all  the  points  of  the  line 
on   the   plane.      Thus   A'B'   is   the 
projection  of  AB  on  the  plane  MN. 


DIHEDKAL    ANGLES 

PROPOSITION  XV.     THEOREM 


339 


552.    Two  dihedral  angles  are  equal  if  their  plane  angle* 
are  equal. 


Given  /.DBF  the  plane  Z  of  the  dihedral  £  C-AB-F, 
£D'B'F'  the  plane  Z  of  the  dihedral  /.C'-A'B'-F',  and 
/.DBF  =  £DfBfFf. 

To  prove          Z  C-AB-F  =  Z  Cf-A'B'-F'. 

Proof.    Apply  the  dihedral  Z  Cf-A'B'-F'  to  Z  C-AB-F 

so  that  /.D'B'F'  coincides  with  its  equal,  /.DBF. 

Greom.  Ax.  2. 

Then  line  A'Bf  must  coincide  with  AB,  Art.  515. 

(/or  A'B'  and  AB  are  loth  ±  plane  DBF  at  the  point  B). 

Hence  the  plane  A'B'Df  will  coincide  with  plane  ABD, 

Art.  501. 
(through  two  intersecting  lines  only  one  plane  can  be  passed). 

Also  the  plane  AfB'Ff  will  coincide  with  the  plane  ABF, 
(same  reason). 

.'.  Z  G'-A'B'-F'  coincides  with  Z  C-AB-F  and  is  equal 

to  it.  Art.  47. 

Q.  E.  D. 

553.  COR.  The  vertical  dihedral  angles  formed  by  two 
intersecting  planes  are  equal. 

In  like  manner,  many  other  properties  of  plane  angles 
are  true  of  dihedral  angles. 


340     '  BOOK   VI.       SOLID     GEOMETRY 

V  PROPOSITION  XVI.     THEOREM 

554.    Two  dihedral  angles  have  the  same  ratio  as  their 
plane  angles. 


Fig.l 


Fig,2 


Jig.  3 


Given  the  dihedral  A  C-AB-D  and  C'-A'B'-D'  having 
the  plane  A  CAD  and  C'A'D',  respectively. 


To    prove    Z  C'-A'B'-D' 
/.CAD. 


Z  C-AB-D   =  Z  C'A'D' 


CASE  I.  When  the  plane  A  C'A'D'  and  CAD  (Figs.  2 
and  1),  are  commensurable. 

Proof.  Find  a  common  measure  of  the  A  C'A'D'  and 
CAD,  as  Z  CAR,  and  let  it  be  contained  in  Z  C'A'D'  n 
times,  and  in  Z  CAD  m  times. 

Then  Z  C'A'D'  :  Z  CAD  =  n  :  m. 

Through  A'B'  and  the  lines  of  division  of  Z  C'A'D'  pass 
planes,  and  through  AB  and  the  lines  of  division  of 
Z  CAD  pass  planes.  These  planes  will  divide  the  dihedral 
Z  C'-A'B'-D'  into  n,  and  /.C-AB-D  into  m  parts,  all 
equal.  Art.  552. 

/.   /.C'-A'B'-D'  :   /.C-AB-D  =  n  :  m. 

Hence  Z  C'-A'B'-D'  :  Z  (J-AB-D  =  Z  C'A'D' :  Z  CAD. 

(Why?) 


DIHEDKAL     ANGLES  341 

CASE  II.  When  the  plane  angles  C'A'D'  and  CAD  (Figs. 
3  and  1)  are  incommensurable. 

Proof.  Divide  the  Z  CAD  into  any  number  of  equal 
parts,  and  apply  one  of  these  parts  to  the  Z  C'A'D1.  It 
will  be  contained  a  certain  number  of  times  with  a  remain- 
der, as  £LA'D',  less  than  the  unit  of  measure. 

Hence  the  A  G'A'L  and  CAD  are  commensurable. 

/.  Z  C'-A'B'-L\tC-AB-D=.  Z  C'A'L:Z.CAD.  Case  I- 

If  now  we  let  the  unit  of  measure  be  indefinitely  dimin- 
ished, the  /.LA'D',  which  is  less  than  the  unit  of  measure, 
will  be  indefinitely  diminished. 

.-.  Z  C'A'L  =  Z  C'A'D'  as  a  limit,  and 

Z  G'—A'B'—L  =  Z  C'—A'B'—D'  as  a  limit.      Art.  251. 

Z  C'  —  A'Bf  —  L 
Hence     —  —  —  —  —  —      becomes     a     variable,      with 

Z.  O~-A-D  ~  JJ 


Z  C'-A'B'-D'       ..    ..    ., 

—  .  as  its  limit;  Art.  253,  3. 

Z.  0     ^4,-D     JL/ 


,  Z 

Also      .  ~  .  j    becomes  a  variable  with  as    its 

limit.  Art.  253,  3. 

tO-A'B'-L  Z  C'A'L 

But  the  variable       .         .  =  the  variable 


Z  \jA.jJ 

always.  Case  I. 


,  Z  C'-A'B'-D'     .,  .   Z 

•"•  the  hmit  =the  hmlt 


Q.  E.  D. 


Ex.  1.  How  many  straight  lines  are  necessary  to  indicate  a  dihe- 
dral angle  (as  /.E-AB-D,  p.  338)?  How  many  straight  lines  are 
necessary  to  indicate  the  plane  angle  of  a  dihedral  angle  ?  Hence, 
what  is  the  advantage  of  using  a  plane  angle  of  a  dihedral  angle 
instead  of  the  dihedral  angle  itself  ? 

Ex.  2.  Give  three  additional  properties  of  dihedral  angles  anal- 
ogous to  properties  of  plane  angles  given  in  Book  I, 


342 


X 


BOOK   VI.       SOLID     GEOMETKY 

PROPOSITION  XVII.     THEOREM 


555.    If  a  straight  line  is  perpendicular  to  a  plane,  every 
plane  draivn  through  that  line  is  perpendicular  to  the  plane. 


Given   the  line  AB   _L   plane  MN,  and   the  plane  PQ 
passing  through  AB  and  intersecting  MN  in  RQ. 

To  prove  PQ  J_  MN. 

Proof.   In  the  plane  MN  draw  BG  JL  RQ  at  B. 

But  AB  _L  RQ.  Art.  505. 

.*.  Z  ABC  is  the  plane  Z  of  the  dihedral  /iP-RQ-M. 

Art.  548. 

But  Z  ABC  is  a  right  Z  ,  Art.  505. 

( for  AB  J.  MN  by  hyp . ) . 


/.  PQ  -L  MN. 


Art.  547. 
Q.  E.  B. 


556.  COR.  A  plane  perpendicular  to  the  edge  of  a 
dihedral  angle  is  perpendicular  to  each  of  the  two  faces  form- 
ing the  dihedral  angle. 


DIHEDEAL    ANGLES  343 

PROPOSITION  XVIII.     THEOREM 

557.  If  two  planes  are  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them  perpendicular  to  their 
line  of  intersection  is  perpendicular  to  the  other  plane. 

p 


Given  the  plane  PQ  JL  plane  MN  and  intersecting  it  in 
the  line  RQ-,   and  AB  a  line  in  PQ  _L  RQ. 
To  prove  AB  J_  plane  MN. 

Proof.    In  the  plane  MN  draw  BC  JL  RQ. 
:.  ZABCis  the  plane  Z  of  the  dihedral  ^P-RQ-M. 

Art.  548. 

/.   Z  ABC  is  a  rt.  Z ,  Art.  554. 

(for  P-RQ-Mis  a  right  dihedral  Z  ). 

.*.  AB  JL  BC  and  RQ  at  their  intersection. 

/.  AB  _L  plane  NN.  (Why  T) 

Q.  £.  D. 

558.  COR.  1.    If  two  planes  are  perpendicular  to  each 
other,  a  perpendicular  to  one  of  them  at  any  point  of  their 
intersection  will  lie  in  the  other  plane. 

For,  in  the  above  figure,  a  -L  erected  at  the  point  B  in 
the  plane  MN  must  coincide  with  AB  lying  in  the  plane 
PQ  and  _L  MN,  for  at  a  given  point  in  a  plane  only  one  JL 
can  be  drawn  to  that  plane  (Art.  515). 

559.  COR.  2.    If  two  planes  are  perpendicular  to  each 
other,  a  perpendicular  to  one  plane,  from  a  point  in  the  other 
plane,  will  lie  in  tlie  other  plane. 


344 


BOOK  VI.      SOLID     GEOMETRY 


PROPOSITION  XIX.     THEOREM 


560.  If  two  intersecting  planes  are  each  perpendicular 
to  a  third  plane.,  their  line  of  intersection  is  perpendicular 
to  the  third  plane. 


Given  the  planes  PQ  and  R8  JL  plane  MN,  and  inter- 
secting in  the  line  AB. 


To  prove 


AB  JL  plane  MN. 


Proof.  At  the  point  B  in  which  the  three  planes  meet 
erect  a  _L  to  the  plane  MN.  This  _L  must  lie  in  the  plane  PQ, 
and  also  in  the  plane  R8.  Art-  558- 

Hence  this  _L  must  coincide  with  AB,  the  intersection 
of  PQ  and  RS.  Art.  508,  2. 

/.  AB  J_  plane  MN. 

Q.  £.  D. 

561.  COR.  If  two  planes,  including  a  right  dihedral 
angle,  are  each  perpendicular  to  a  third  plane,  the  intersec- 
tion of  any  two  of  the  planes  is  perpendicular  to  the  third 
plane,  and  each  of  the  three  lines  of  intersection  is  perpen' 
dicular  to  the  other  two. 


Ex.  1.    Name  all  the  dihedral  angles  on  the  above  figure. 
Ex.  2.    If  Z  CBQ=3Q°,  find  the  ratio  of  each  pair  of  dihedral 


DIHEDRAL    ANGLES 


345 


PROPOSITION  XX.     THEOREM 

562.  Every  point  in  the  plane  which  bisects  a  given 
dihedral  angle  is  equidistant  from  the  faces  of  the  dihedral 
angle. 


Given  plane  CB  bisecting  the  dihedral  /.A-BR-D,  P 
any  point  in  plane  BC,  PQ  and  PT  JL  faces  BA  and  BD, 
respectively. 

To  prove  PQ  =  PT. 

Proof.  Through  PQ  and  PT  pass  a  plane  intersecting 
AB  in  QR,  BD  in  JBT,and  BG  in  PR. 

Then  plane  PQT  JL  planes  AB  and  £D.  Art.  555. 

/.  plane  PQT  J_  line  RB,  the  intersection  of  the  planes 

AB  and  BD.  Art.  560. 

/.  I2£  J_  RQ,  RP  and  -RT.  Art.  505. 

/.  A  QRP  and  PRT  are  the  plane  A  of  the  dihedral 
^  A-BR-P  and  P-BR-D.  Art.  548. 

But  these  dihedral  ^  are  equal.  Hyp. 

Art.  554. 
(Why?) 

(Why?) 
Q.  £.  D. 


rt.A  PQR=rt.  A  PRT. 
:.  PQ=PT. 


563.    The  locus  of  all  points  equidistant  from  the  faces 
of  a  dihedral  angle  is  the  plane  bisecting  the  dihedral  angle. 


346 


BOOK   VI.       SOLID     GEOMETRY 


PROPOSITION  XXI.     PROBLEM 

564.    Through  any  straight  line  not  perpendicular  to  a 
given  plane,  to  pass  a  plane  perpendicular  to  the  given  plane. 


•if 


Given  the  line  AB  not  J_  plane  MN. 

To  construct  a  plane  passing  through  AB  and  _L  MN. 

Construction.  From  a  point  A  in  the  line  AB  draw  a  _L 
AC  to  the  plane  MN.  Art.  516. 

Through  the  intersecting  lines  AB  and  AC  pass  the 
plane  AD.  Art.  503. 

Then  AD  is  the  plane  required. 

Proof.    The  plane  AD  passes  through  AB.  Constr. 

Also  plane  AD  _L  plane  MN,  Art.  555. 

(for  it  contains  AC,  which  is  _L  MN). 

Q.  E.  F. 

565.  COR.  1.    Through  a  straight  line  not  perpendicular 
to  a  given  plane  only  one  plane  can  be  passed  perpendicu- 
lar to  that  plane. 

For,  if  two  planes  could  be  passed  through  AB  _L  plane 
MN,  this  intersection  AB  would  be  J_  MN  (Art.  560), 
which  is  contrary  to  the  hypothesis. 

566.  COR.  2.    The  projection  upon  a  plane  of  a  straight 
line  not  perpendicular  to  that  plane  is  a  straight  line. 

For,  if  a  plane  be  passed  through  the  given  line  J_  to 
the  given  plane,  the  foot  of  a  _L  from  any  point  in  the 
line  to  the  given  plane  will  be  in  the  intersection  of  the 
two  planes  (Art.  559). 


DIHEDRAL    ANGLES  347 

PROPOSITION  XXII.     THEOREM 

567.  The  acute  angle  which  a  line  makes  with  its  pro- 
jection on  a  plane  is  the  least  angle  which  it  makes  with  any 
line  of  the  plane  through  its  foot. 


Given  line  AB  meeting  the  plane  MN  in  the  point  B, 
BG  the  projection  of  AB  on  MN,  and  PB  any  other  line 
in  the  plane  MN  through  B. 

To  prove  that  Z  ABC  is  less  than  Z  ABP. 

Proof.    Lay  off  PB  equal  to  CB,  and  draw  AC  and  AP. 

Then,  in  the  A  ABC  and  ABP, 

AB=AB.  (Why?) 

BC=BP.  (Why!) 

But  AC  <  AP.  Art.  521. 

/.  Z  ABC  is  less  than  Z  ABP  Art.  108. 

Q.  £.  D. 

568.    DEF.    The  inclination  of  a  line  to  a  plane  is  the 

acute  angle  which  the  given  line  makes  with  its  projection 
upon  the  given  plane. 


Ex.  1.  A  plane  has  an  inclination  of  47°  to  each  of  the  faces  of  a 
dihedral  angle  and  is  parallel  to  the  edge  of  the  dihedral  angle;  how 
many  degrees  are  in  the  plane  angle  of  the  dihedral  angle? 

Ex.  2.  In  the  figure  on  page  345,  if  PT=QT,how  large  is  the 
dihedral  L  A-BR-Dt  if  PT—RT.  how  large  is  it? 


348 


BOOK   VI.       SOLID     GEOMETKY 


PROPOSITION  XXIII.    PROBLEM 

569.    To  draw  a  common  perpendicular  to  any  two  lines 
<ot  in  the  same  plane, 
a         L 


Given  the  lines  AB  and  CD  not  in  the  same  plane. ' 
To  construct  a  line  perpendicular  to  both  AB  and  CD. 
Construction.    Through  AB  pass  a  plane  MN\\  line  CD. 

Art.  529. 

Through  CD  pass  a  plane  CF  J_  plane  MN  (Kii.  564),  and 
intersecting  plane  MN  in  the  line  EF. 

Then  EF  \\  CD  (Art.  528),  /.  EF  must  intersect  AB 
(which  is  not  ||  CD  by  hyp.)  in  some  point  K. 

At  K  in  the  plane  CF  draw  LK  J_  EF.  Art,  274. 

Then  LK  is  the  perpendicular  required. 

Proof.  LK  _L  EF.  Constr. 

.*.  LK  J_   CD.  Art.  123. 

Also                         LK  _L  plane  MN.  Art.  557. 

/.  LK  _L  line  AS.  (Why  ?) 
/.  ££:  -L  both  CD  and  AS. 

<?.  E.  F. 

570.  Only  one  perpendicular  can  be  drawn  between  two 
lines  not  in  the  same  plane. 

For,  if  possible,  in  the  above  figure  let  another  line  BD 
be  drawn  _L  AB  and  CD.  Then,  if  a  line  be  drawn  through 
B  ||  CD,  BD  _L  this  line  (Art.  123),  and  .'.  J_  plane  MN 
(Art.  509).  Draw  DF  JL  line  EF;  then  DF  J_  plane  MN 
(Art.  557) .  Hence  from  the  point  D  two  _k  ,  DB  and  DF, 
are  drawn  to  the  plane  MN,  which  is  impossible  (Art.  517). 


POLYHEDRAL    ANGLES 


349 


POLYHEDRAL   ANGLES 

571.  A  polyhedral  angle  is  the  amount 
of  opening  between  three  or  more  planes 
meeting  at  a  point. 

Such  an  angle  may  be  regarded  as  a  portion  of 
space  cut  out  by  the  planes  forming  the  angle. 

572.  The  vertex  of  a  polyhedral  angle 

is  the  point  in  which  the  planes  forming  the  angle  meet;  the 
edges  are  the  lines  in  which  the  planes  intersect;  the  faces 
tore  the  portions  of  the  planes  forming  the  polyhedral  angle 
which  are  included  between  the  edges ;  the  face  angles  are 
the  angles  formed  by  the  edges. 

Each  two  adjacent  faces  of  a  polyhedral  angle  form  a 
dihedral  angle. 

The  parts  of  a  polyhedral  angle  are  its  face  angles  and 
dihedral  angles  taken  together. 

573.  Naming  a  polyhedral  angle.     A  polyhedral  angle 
is  named  either  by  naming  the  vertex,  as  F;   or  by  naming 
the  vertex  and  a  point  on  each  edge,  as  V-ABC. 

In  case  two  or  more  polyhedral  angles  have  the  same 
vertex,  the  latter  method  is  necessary. 

In  the  above  polyhedral  angle,   the  vertex  is  V;    the 
edges  are  VA,  VB,  VC-,   the  face  angles 
are  AVB,  BVC,  AVC. 

574.  A  convex  polyhedral  angle  is  a 
polyhedral    angle    in    which    a    section 
made  by  a  plane  cutting  all  the  edges 
is  a  convex  polygon,  as  V-ABCDE. 

575.  A  trihedral  angle  is  a  polyhe- 
dral angle  having  three  faces;    a  tetrahedral  angle  is  one 
having  four  faces,  etc. 


350 


BOOK  VI,       SOLID     GEOMETRY 


576.  A  trihedral  angle  is  rectangular,  birectangular,  or 
trirectangular,  according  as  it  contains  one,  two,  or  three 
right  dihedral  angles. 

577.  An  isosceles  trihedral  angle  is  a   trihedral  angle 
two  of  whose  face  angles  are  equal. 


578.  Vertical  polyhedral  angles  are  polyhedral  angles 
having  the  same  vertex  and  the  faces  of  one  the  faces  of 
the  other  produced. 


579.  Two   equal    polyhe- 
dral angles  are  polyhedral  an- 
gles having  their  correspond- 
ing parts  equal  and  arranged 
in  the  same  order,  as  V-ABG 
and  V'-A'B'C1. 

Two  equal  polyhedral  angles 
may  be  made  to  coincide.    . 

580.  Two  symmetrical  poly- 
hedral angles  are  polyhedral  an- 
gles having  their  corresponding 
parts    equal    but    arranged    in 
reverse  order. 


If  the  faces  of  a  trihedral  angle,  F-ABC,  be 
produced,  they  will  form  a  vertical  trihedral  angle, 
V-A'B'C',  which  is  symmetrical  to  V-ABC.  For, 
if  V-A'B'C'  be  rotated  forward  about  a  horizontal 
axis  through  V,  the  two  trihedral  angles  are  seen 
to  have  their  corresponding  parts  equal  but  ar- 
ranged in  reverse  order. 

Similarly,  any  two  vertical  polyhedral  angles 
are  symmetrical. 


CL-l 


POLYHEDRAL    ANGLES  351 

581.  Equivalence  of  symmetrical  polyhedral  angles.  It 
has  been  shown  in  Plane  Geometry  (Art.  488)  that  two 
triangles  (or  polygons)  symmetrical  with  respect  to  an 
axis  have  their  corresponding  parts  equal  and  arranged 
in  reverse  order.  By  sliding 
two  such  figures  about  in  a  plane 
they  cannot  be  made  to  coincide, 

t 

but  by  lifting  one  of  them  up 

from  the  plane  in  which  it  lies  and  turning  it  over  it  may 

be  made  to  coincide  with  the  other  figure. 

Symmetrical  polyhedral  angles,  however,  cannot  be 
made  to  coincide  in  any  way;  hence  some  indirect  method 
of  showing  their  equivalence  is  necessary.  See  Ex.  29, 
p.  358,  and  Arts.  789-792. 


Ex.  1.    Name  the  trihedral  angles  on  the  figure  to  Prop.  XX.     If 
=>90°,  what  kind  of  trihedral  angles  are  those  on  the  figure  ? 
If  ZPJ2$=30°,  what  kind  are  they? 

Ex.  2.  Are  two  trirectangular  trihedral  angles  necessarily  equal  ? 
Prove  this. 

Ex.  3.  Are  two  lines  which  are  perpendicular  to  the  same  plane 
necessarily  parallel  ?  Are  two  planes  which  are  perpendicular  to  the 
same  plane  necessarily  parallel  ?  Are  two  planes  which  are  perpen- 
dicular to  the  same  line  necessarily  parallel  ? 

Ex.  4.  Let  the  pupil  cut  out  three  pieces  of  pasteboard  of  the  form 
indicated  in  the  accompanying  figures ;  cut  them  half  through  where 
the  lines  are  dotted ;  fold  them  and  fasten  the  edges  so  as  to  form 
three  trihedral  angles,  two  of  which  (Figs.  1  and  2)  shall  be  equal 
and  two  (Figs.  1  and  3)  symmetrical.  By  experiment,  let  the  pupil 
find  which  pair  may  be  made  to  coincide,  and  which  not.  | 


Fig. 


352 


BOOK   VI.       SOLID     GEOMETKY 


PROPOSITION  XXIV.     THEOREM 

582.    The  sum  of  any  two  face  angles   of  a    trihedral 
angle  is  greater  than  the  third  face  angle. 


Given  the  trihedral  angle  S-ABC,  with  angle  A8C  its 
greatest  face  angle. 

To  prove  Z  ASB  +  ^BSC  greater  than  Z  A8C. 
Proof.    In  the   face  ASC  draw  SD,   making  ^ASD= 
/.ASB. 

Take  SD  =  SB. 

In  the  face  ASC  draw  the  line  ADC  in  any  convenient 
direction,  and  draw  AB  and  BC. 

Then,  in  the  A  ASB  and  ASD,  SA  =  SA. 
SB  =  SD,  and  /.ASB=  Z.ASD. 
•  .'.  A  ASB=  A  ASD. 

/.  AB  =  AD. 

Also  AB  +  BC  >  AC. 

Hence,  subtracting  the  equals  AB  and 

BC  >  DC. 

Hence,  in  the  A  D£C  and  DSC,  SC=SC, 
>  DC. 

/.   Z  BSC  is  greater  than  /.DSC. 


(Why?) 
(Why?) 
(Why?) 
(Why?) 
(Why  ?) 

(Why?) 

=  SD,  and 

(Why?) 
Art.  108. 


To  each  of  these  unequals  add  the  equals  /.ASB  and 


:.   /.ASB+  Z  BSC  is  greater  than  /.ASC.    (Why?) 

Q.  E.  D. 

Ex.    In  the  above  figure,   if   /.ASC  equals  one  of  the  other   face 
angles  at  5,  as  Z  ASB,  how  is  the  theorem  proved  ? 


POLYHEDRAL    ANGLES  353 


PROPOSITION  XXV.     THEOREM 

i 

583.    The  sum  of  the  face  angles  of  any  convex  polyhedral 
angle  is  less  than  four  right  angles. 


Given  the  polyhedral  angle  S-ABCDE. 

To  prove  the  sum  of  the  face  A  at  8  less  than  4  rt.  A  . 

Proof.  Pass  a  plane  cutting  the  edges  of  the  given  poly- 
hedral angle  in  the  points  A,  B,  C,  D,  E. 

From  any  point  O  in  the  polygon  ABCDE  draw  OA 
OB,  OC,  OD,  OE. 

Denote  the  A  having  the  common  vertex  S  as  the  8  A, 
and  those  having  the  common  vertex  0  as  the  0  A . 

Then  the  sum  of  the  A  of  the  8  A  =  the  sum  of  A  of 
the  0  A .  Art.  134. 

But  Z  SB  A  +  Z  SBC  is  greater  than  Z  ABC, 


Z£C£+Z  SCD  is  greater  than  ^BCD,  etc/Art 
.*.  the  sum  of  the  base  A  of  the  S  A  >  the  sum  of  the 
base  A  of  the  0  A.  Ax.  9. 

.  •.  the  sum  of  the  vertex  A  of  the  8  A  <  the  sum  of 
the  vertex  A  of  the  0  A,  Ax.  11. 

(if  unequals  be  subtracted  from  equals,  the  remainders  are  unequal 
in  reverse  order). 

But  the  sum  of  the  A  at  0  =  4  rt.   A  .  (Why?) 

.%  the  sum  of  face  A  at  S  <  4  rt.  A  .  Ax.  8. 

Q.  E.  D. 

W 


354  BOOK  VI.       SOLID     GEOMETRY 

PROPOSITION  XXVI.     THEOREM 

584.  //  two  trihedral  angles  have  the  three  face  angles 
of  one  equal  to  the  three  face  angles  of  the  other,  the  tri- 
hedral angles  have  their  corresponding  dihedral  angles  equal, 
and  are  either  equal  or  symmetrical,  according  as  their 
corresponding  face  angles  are  arranged  in  the  same  or  in 
reverse  order. 


Given  the  trihedral  A  S-ABC  and  S'-A'B'C',  having 
the  face  A  ASB,  A  SO  and  BSC  equal  to  the  face  A  A'S'B', 
A'S'C'  and  B'S'C',  respectively. 

To  prove  that  the  corresponding  dihedral  A  of  8- ABC 
and  S'-A'B'C'  are  equal,  and  that  A  S-ABC  and  S'- 
A'B'C'  are  either  equal  or  symmetrical. 

Proof.  On  the  edges  of  the  trihedral  A  take  8A,  SB, 
SO,  S'A',  S'B',  S'C'  all  equal. 

Draw  AB,  AC,  BC,  A'B',  A'C',  B'C'. 

Then,  1.  In  the  A  ASB  and  A'S'B',  SA  =  S'Af,  SB  = 
S'B',  and  /iASB  =  /.A'S'B'.  (Why?) 

/.  A  A8B  =  A  A'8'B'.  (Why?) 

.'.  AB=A'B'.  (Why?) 
2     In  like  manner      AC=A'Cf,  and  BC=B'C'. 

.'.  A  ABC=&  A' B'C'.  (Why?) 


EXERCISES     ON     THE     LINE     AND    PLANE  355 

3.  Take  D  a  convenient  point  in  SA,  and  draw  DE  in 
the  face  ASB,  and  DF  in  the  face  ASC,  each  J_  8A. 

DE  and  DF  meet  AB  and  A  G  in  points  E  and   F, 

T"P^T)Pf*f  1  VP  1  V 

(/or  /£  SAB  and  SAC  are  acute). 

Similarly,  take  SfDf  =  SD  and  construct  A  D'E'F'. 

Then,  in  the  rt.  A  ADE  and  A'D'E',  AD=A'D',  and 

/.DAE^D'A'E'.  (Why  ?) 

.-.  A  ADE=&  A'D'E'.  (Why?) 

/.  AE=A'Ef,  and  DE=D'Ef.  (Why?) 

4.  In  like  manner  it  maybe  shown  that  AF=A'F'1  and 
DF=D'F'. 

:.  A  AEF=&  A'E'F'.  (Why?) 

And  ^7jP=  JEJ'JF'  (Why  ?) 


5.    Hence,  in  the  A  DEF  and  D'^J^P,  DE=D'E',  DF 

=  DfFf  and  ^7^=  E'F'.  (Why  ?) 

/.  A  DEF=&  D'E'F'.  (Why?) 

.*.   Z  EDF=  Z  ^D^.  (Why  ?) 

But  these  A  are  the  plane  A  of  the  dihedral  A  whose 

edges  are  SA  and  S1  'A'. 

:.  dihedral  Z  B-A8~C=  dihedral  £B'-A'8'-C?    Art.  552. 
In  like  manner  it  may  be  shown  that  the  dihedral  A  at 
SB  and  S'B'  are  equal;   and  that  those  at  8C  and  S'G'  are 
equal. 

.'.  the  trihedral  A  S  and  Sf  are  either  equal  or  sym- 
metrical. Arts.  579,  580. 

Q.  E.  D. 

EXERCISES.     CROUP  64 

THEOREMS   CONCERNING   THE   LINE   AND   PLANE   IN   SPACE 

Ex.  1.    A  segment  of  a  line  not  parallel  to  a  plane  is  longer  than 
its  projection  in  the  plane. 

Ex.  2.    Equal    straight  lines  drawn  from  a  point  to   a   plane  are 
equally  inclined  to  the  plane, 


356 


BOOK   VI.       SOLID     GEOMETRY 


B 


N 


Ex.  3.  A  line  and  plane  perpendicular  to  the  same  plane  are 
parallel. 

Ex.  4.  If  three  planes  intersecting  in  three  straight  lines  are 
perpendicular  to  a  plane,  their  lines  of  intersection  are  parallel. 

Ex.  5.  If  a  plane  bisects  any  line  at  right 
angles,  any  point  in  the  plane  is  equidistant 
from  the  ends  of  the  line. 

Ex.  6.  Given  AB  _L  plane  MN, 
and  AC  J_  plane  BS; 
prove  BC  _L  NE. 

Ex.  7.    Given  PQ  J_  plane  MN, 
PR  _L  plane  BL, 
and     BS  J_  plane  .MN; 
prove  QS  I.  AB.  M 

Ex.  8.  If  a  line  is  perpendicular  to  one 
of  two  intersecting  planes,  its  projection 
on  the  other  plane  is  perpendicular  to  the 
line  of  intersection  of  the  two  planes. 

Ex.  9.    Given  CE  _L  DE, 
AE  J_  DE, 

and     /  C-AD-E  a  rt.  dihedral  Z  ; 
prove  CA  J_  plane  DAE. 

Ex.  10.  The  projections  of  two  parallel  lines  on 
a  plane  are  parallel.  (Is  the  converse  of  this 
theorem  also  true  ?) 

Ex.  11.  If  two  parallel  planes  are  cut  by  two  non -parallel  planes, 
the  two  lines  of  intersection  in  each  of  the  parallel  planes  will  make 
equal  angles. 

Ex.  12.  If  a  line  is  perpendicular  to  a  plane,  any  plane  parallel 
to  the  line  is  perpendicular  to  the  plane.  (Is  the  converse  true  ?) 

Ex.  13.  In  the  figure  to  Prop.  VI,  given  AB  J_  MN  and  AF  J. 
DC;  prove  BF  J_  DC. 

Ex.  14.  Two  planes  parallel  to  a  third  plane  are  parallel  to  each 
other. 

[Sue.     Draw  a  line  -L  third  plane.] 


EXERCISES  ON  THE  LINE  AND  PLANE     357 

Ex.  15.  The  projections  upon  a  plane  of  two  equal  and  parallel 
straight  lines  are  equal  and  parallel. 

Ex.  16.  A  line  parallel  to  two  planes  is  parallel  to  their  inter- 
section. 

Ex.  17.  In  the  figure  to  Prop.  XXII,  if  angle  GBP  is  obtuse, 
prove  the  angle  ABP  obtuse. 

Ex.  18.  In  a  quadrilateral  in  space  (i.  e.,  a 
quadrilateral  whose  vertices  are  not  all  in  the 
same  plane),  show  that  the  lines  joining  the 
midpoints  of  the  sides  form  a  parallelogram. 

Ex.  19.  The  lines  joining  the  midpoints  of 
the  opposite  sides  of  a  quadrilateral  in  space  bisect  each  other. 

Ex.  20.  The  planes  bisecting  the  dihedral  angles  of  a  trihedral 
angle  meet  in  a  line  every  point  of  which  is  equidistant  from  the 
three  faces. 

[SuG.     See  Art.  562.] 

Ex.  21.    Given  OQ  bisecting 

PQ  JL  plane  ROS, 
QR  L  OR, 
and      QS  _L  OS', 
prove  PR=  PS,  PR  JL  OR, 
and  PS  _L  OS. 

Ex.  22.  In  a  plane  bisecting  a  given  plane  angle,  and  perpendicu- 
lar to  its  plane,  every  point  is  equidistant  from  the  sides  of  the  angle. 

[SuG.  See  Ex.  21;  or  through  P  any  point  in  the  bisecting  plane 
pass  planes  -L  to  the  sides  of  the  Z  ,  etc.] 

Ex.  23.  In  a  trihedral  angle,  the  three  planes  bisecting  the  three 
face  angles  at  right  angles  to  their  respective  planes,  intersect  in  a  line 
every  point  of  which  is  equidistant  from  the  three  edges  of  the  tri- 
hedral angle. 

Ex.  24.  If  two  face  angles  of  a  trihedral  angle  are  equal,  the  dihe- 
dral angles  opposite  them  are  equal. 

Ex.  25.  In  the  figure  to  Prop.  XXIV,  prove  that  £ASC+  £BSC 
is  greater  than  Z  ASD  +  Z  BSD. 

Ex.  26.  The  common  perpendicular  to  two  lines  in  space  is  the 
shortest  line  between  them. 


358 


BOOK   VI.       SOLID     GEOMETKY 


Ex.  27.    Given      MN  \\  ES, 
and  PB  =  pi ; 

prove  /.ABC=/.abc, 
and  A   ABC  =  A  abc. 


Ex.  28.    Two    isosceles 
trihedral  angles  are  equal. 


symmetrical 


r\ 


\ 


Ex.  29.    Any  two  symmetrical  trihe- 
dral angles  are  equivalent. 

[SUG.  Take  SA,  SB,  SC,  S'A', 
S'B',  S'C',  all  equal.  Pass  planes  ABC, 
A'B'C'.  Draw  SO  and  S'O'  ±  these 
planes.  Then  the  trihedral  A  are 
divided  into  three  pairs  of  isosceles 
symmetrical  trihedral  A,  etc.] 


EXERCISES.     CROUP  65 

LOCI   IN   SPACE 
Find  the  locus  of  a  point  equidistant  from 

Ex.  1.    Two  parallel  planes.  Ex.  3.  Three  given  points. 

Ex.  2.    Two  given  points.  Ex.  4.    Two  intersecting  lines. 

Ex.  5.    The  three  faces  of  a  trihedral  angle. 
Ex.  6.    The  three  edges  of  a  trihedral  angle. 

Find  the  locus 

Ex.  7.  Of  all  lines  passing  through  a  given  point  and  parallel  to 
a  given  plane. 

Ex.  8.  Of  all  lines  perpendicular  to  a  given  line  at  a  given  "point 
in  the  line. 

Ex.  9.  Of  all  points  in  a  given  plane  equidistant  from  a  given 
point  outside  the  plane. 

Ex.  10.  Of  all  points  equidistant  from  two  given  points  and  from 
two  parallel  planes. 

Ex.  11.  Of  all  points  equidistant  from  two  given  points  and  from 
two  intersecting  planes. 

Ex.  12.  Of  all  points  at  a  given  distance  from  a  given  plane  and 
equidistant  from  two  intersecting  lines. 


EXERCISES     ON     THE     LINE     AND     PLANE  359 

EXERCISES.     CROUP  66 

PROBLEMS   CONCERNING   THE   POINT,    LINE   AND   PLANE 
IN   SPACE 

Ex.  1.  Through  a  given  point  pass  a  plane  parallel  to  a  given 
plane. 

Ex.  2.  Through  a  given  point  pass  a  plane  perpendicular  to  a 
given  plane. 

Ex.  3.  Through  a  given  point  to  construct  a  plane  parallel  to  two 
given  lines  which  are  not  in  the  same  plane. 

Prove  that  only  one  plane  can  be  constructed  fulfilling  the  given 
conditions. 

Ex.  4.    Bisect  a  given  dihedral  angle. 

Ex.  5.  Draw  a  plane  equally  inclined  to  three  lines  which  meet  at 
a  point. 

Ex.  6.  Through  a  given  point  draw  a  line  parallel  to  two  given 
intersecting  planes. 

Ex.  7.    Find  a  point  in  a  plane  such  that  lines  drawn  to  it  from 
two  given  points  without  the  plane  make  equal  angles  with  the  plane. 
[Sua.     See  Ex.  23,  p.  176.] 

Ex.  8.  Find  a  point  in  a  given  line  equidistant  from  two  given 
points. 

Ex.  9.    Find  a  point  in  a  plane  equidistant  from  three  given  points. 

Ex.  10.  Find  a  point  equidistant  from  four  given  points  not  in  a 
plane. 

Ex.  11.  Through  a  given  point  draw  a  line  which  shall  intersect 
two  given  lines. 

[SUG.  Pass  a  plane  through  the  given  point  and  one  of  the  given 
lines,  and  pass  another  plane  through  the  given  point  and  the  other 
given  line,  etc.] 

Ex.  12.  Through  a  given  point  pass  a  plane  cutting  the  edges  of 
a  tetrahedral  angle  so  that  the  section  shall  be  a  parallelogram. 

[SuG.  Produce  each  pair  of  opposite  faces  to  intersect  in  a 
Straight  line,  etc.] 


BOOK  VII 


Polyhedron 


POLYHEDRONS 

585.  A  polyhedron   is   a    solid    bounded    by 

586.  The  faces   of  a  polyhe- 
dron are  its  bounding  planes;   the 
edges  of  a  polyhedron  are  the  lines 
of  intersection  of  its  faces. 

A  diagonal  of  a  polyhedron  is 
a  straight  line  joining  two  of  its 
vertices  which  are  not  in  the 
same  face.  The  vertices  of  a  poly- 
hedron are  the  points  in  which  its 
edges  meet  or  intersect. 

587.  A  convex  polyhedron  is  a  polyhedron  in  which  a 
section  made  by  any  plane  is  a  convex  polygon. 

Only  convex  polyhedrons  are  to  be  considered  in  this  book. 

588.  Classification    of    polyhedrons.     Polyhedrons    are 
sometimes    classified   according    to   the   number   of   their 
faces.     Thus,  a  tetrahedron  is  a  polyhedron  of  four  faces; 
a  hexahedron  is  a  polyhedron  of  six  faces ;   an  octahedron 
is    one  of  eight,   a  dodecahedron  one  of   twelve,   and  an 
icosahedron  one  of  twenty  faces. 


Tetrahedron        Cube 


Octahedron    Dodecahedron      icosahedron 
(360) 


POLYHEDEONS  361 

The  polyhedrons  most  important  in  practical  life  are  those  deter- 
mined by  their  stability,  the  facility  with  which  they  can  be  made  out 
of  common  materials,  as  wood  and  iron,  the  readiness  with  which 
they  can  be  packed  together,  etc.  Thus,  prism  means  "something 
sawed  off." 

PRISMS    AND    PARALLELOPIPEDS 

589.  A  prism  is  a  polyhedron  bounded 
by  two   parallel    planes  and  a  group  of 
planes   whose    lines    of    intersection    are 
parallel, 

590.  The  bases  o$   a   prism   are   the 
faces  formed  by  the  two  parallel  planes; 

Prism 

the  lateral  faces  are  the  faces  formed  by 

the  group  of  planes  whose  lines  of  intersection  are  parallel. 

The  altitude  of  a  prism  is  the  perpendicular  distance 
between  the  planes  of  its  bases. 

The  lateral  area  of  a  prism  is  the  sum  of  the  areas  of  the 
lateral  faces. 

591.  Properties  of  a  prism  inferred  immediately. 

1.  The  lateral  edges  of  a  prism  are  equal,  for  they  are 
parallel  lines  included  between  parallel  planes  (Art.  589) 
and  are  therefore  equal  (Art.  532). 

2.  The  lateral  faces  of  a  prism  are  parallelograms  (Art. 
160),  for  their  sides  formed  by  the  lateral  edges  are  equal 
and  parallel. 

3.  The  bases  of  a  prism  are  equal  polygons,  for  their 
homologous  sides   are  equal  and  parallel,  each   to  each, 
(being  opposite  sides  of  a  parallelogram),  and  their  homol- 
ogous angles  are  equal  (Art.  538). 

592.  A  right  section  of  a  prism  is  a  section  made  by  a 
plane  perpendicular  "to  the  lateral  edges. 


362 


BOOK   VII.       SOLID     GEOMETRY 


593.  A  triangular  prism  is  a  prism  whose  base  is  a 
triangle  ;  a  quadrangular  prism  is  one  whose  base  is 
a  quadrilateral,  etc. 


ObliQiie  Prism? 


Right  Prism 


Regular  Prism 


594.  An  oblique  prism  is  a  prism  whose  lateral  edges 
are  oblique  to  the  bases. 

595.  A  right  prism  is  a  prism  whose 
lateral  edges  are  perpendicular  to  the  bases. 

596.  A  regular  prism  is  a  right  prism 
whose  bases  are  regular  polygons. 

597.  A  truncated  prism  is  that  part  of 
a  prism  included  between  a  base  and  a 
section  made  by  a  plane   eliliqoo»tQ  the 

base  and  cutting  all  the  lateral  edges.  Truncated  Prism 

598.  A  parallelepiped  is  a  prism  whose  bases  are  paral- 
lelograms. 

Hence,  all  the  faces  of  a  parallelepiped  are  parallelograms. 


Oblique 
Parallelepiped 


Right  Rectangular  Cube 

Parallelepiped         Parallelepiped 


599.    A  right  parallelepiped  is  a  parallelepiped  whose 
lateral  edges  are  perpendicular  to  the  bases, 


PRISMS     AND     PAKALLELOPIPEDS  363 

600.  A  rectangular  parallelepiped  is  a  right  parallele- 
piped whose  bases  are  rectangles. 

Hence,  all  the  faces  of  a  rectangular  parallelopiped  are 
rectangles. 

601.  A  cube  is  a  rectangular  parallelopiped  whose  edges 
are  all  equal. 

Hence,  all  the  faces  of  a  cube  are  squares. 

602.  The  unit  of  volume  is  a  cube  whose  edge  is  equal 

to  some  linear  unit,  as  a  cubic  inch,  a  cubic  foot,  etc.  o '*"*{' 

603.  The  volume  of  a  solid  is  the  number  of  units  of 
volume  which  the  solid  contains. 

Being  a  number,  a  volume  may  often  be  determined  from  other 
numbers  in  certain  expeditious  ways,  which  it  is  one  of  the  objects  of 
geometry  to  determine. 

604.  Equivalent   solids  are  solids  whose  volumes   are 
equal. 

Ex.  1.  What  is  the  least  number  of  faces  which  a  polyhedron  can 
have  ? 

Ex.  2.    A  square  right  prism  is  what  kind  of  a  parallelopiped  ? 

Ex.  3.  Are  there  more  right  parallelepipeds  or  rectangular  paral- 
lelepipeds ?  That  is,  which  of  these  includes  the  other  as  a  special 

case  ? 

Ex.  4.  Prove  that  if  a  given  straight  line  is  perpendicular  to  a 
given  plane,  and  another  straight  line  is  perpendicular  to  another 
plane,  and  the  two  planes  are  parallel,  then  the  two  given  lines  are 
^parallel. 


364  BOOK  vn.    SOLID  GEOMETRY 

O  PROPOSITION  I.     THEOREM 

605.    Sections  of  a  prism  made  by  parallel  planes  cutting 
all  the  lateral  edges  are  equal  polygons. 


Given  the  prism  PQ  cut  by  ||  planes  forming  the  sections 
AD  and  A'D' . 

To  prove  section  AD  =  section  A'D' . 

Proof.  AB,  EC,  CD,  etc.,  are  ||  A'B',  B'C',  C'D',  etc., 
respectively.  Art.  531. 

/.  AB,  BC,   <7D?  etc.,   are  equal  to  A'B',  B'C',  C'D'r 

etc.,  respectively.  Art.  157. 

Also  A  ABC,  BCD,  etc.,  are  equal  to  A.  A' B'C',  E'UD', 
etc.,  respectively.  Art. 538. 

/.  ABCDE=A'B'C'D'E',  Art.  47. 

(for  the  polygons  have  all  their  parts  equal,  each  to  each,  and  .'.  can  be 
made  to  coincide). 

Q.  E.  D. 

606.  COR.  1.    Every  section  of  a  prism  made  by  a  plane 
parallel  to  the  base  is  equal  to  the  base. 

607.  COR.  2.    All  right  sections  of  a  prism  are  equal. 


PRISMS  365 


0       PROPOSITION  II.     THEOREM 

608.    The  lateral  area  of  a  prism  is  equal  to  the  product 
of  the  perimeter  of  a  right  section  by  a  lateral  edge. 


Given  the  prism  RQ,  with  its  lateral  area  denoted  by  8 
and  lateral  edge  by  E\  and  AD  a  right  section  of  the  given 
prism  with  its  perimeter  denoted  by  P. 
To  prove  8=PXE. 

Proof.  In  the  prism  RQ,  each  lateral  edge  =  ^7.  Art.  591, 1. 
Also  AB  JL  OR,  BG  JL  IJ,  etc.  Art.  505. 

Hence      area  CU  RH=AB  X  GH=AB  X  E^\ 

area  C3  GJ=  BC  X  E,  \    Art.  385. 

area  EU  IQ  =  CD  X  E,  etc. 

But  S,  the  lateral  area  of  the  prism,  equals  the  sum  of 
the  areas  of  the  ££7  forming  the  lateral  surface. 

/.  adding,  8=  (AB  +  BG  +  CD  +  etc. )  X  E.     Ax.  2. 
Or  S=PXE. 

Q.  £.  D. 

609.    COR.    The  lateral  area  of  a  right  prism  equals  the 
product  of  the  perimeter  of  the  base  by  the  altitude. 


Ex.  Find  the  lateral  area  of  a  right  prism  whose  altitude  is  12  in., 
and  whose  base  is  an  equilateral  triangle  with  a  side  of  6  in.  Also 
find  the  total  area  of  this  figure. 


366  BOOK  VII.      SOLID    GEOMETRY 

O         PROPOSITION  III.     THEOREM 
i 

610.  If  two  prisms  have  the  three  faces  including  a 
trihedral  angle  of  one  equal,  respectively,  to  the  three  faces 
including  a  trihedral  angle  of  the  other,  and  similarly 
placed,  the  prisms  are  equal. 


Given  the  prisms  AJ  and  AfJf,  having  the  faces  AK, 
AD,  AG  equal  to  the  faces  A'K',  A'D',  A'O',  respectively, 
and  similarly  placed. 

To  prove  AJ=A'J'. 

Proof.   The  face    A    EAF,  EAB  and  EAF  are  equal, 

respectively,  to  the  face  A  E'A'F',  EfA'Bf  and  B'A'F'.    Hyp. 

/.  trihedral  ZA  =  trihedral  /.A'  '  Art.  584. 

Apply  the  prism  A'J'  to  the  prism  AJ,  making  each  of 
the  faces  of  the  trihedral  Z  Af  coincide  with  corresponding 
equal  face  of  the  trihedral  /.A.  Geom.  Ax.  2. 

.*.  the   plane  F'J'  will   coincide   in    position    with    the 

plane  FJ,  Art.  500. 

(for  the  points  Gf,  Ff,  K'  coincide  with  G,  F,  K,  respectively). 

Also  the  point  Cr  will  coincide  with  the  point  C. 

:.  G'R'  will  take  the  direction  of  CH.    Geom.  Ax.  3. 
.'.  R'  will  coincide  with  H.  Art.  508,  1. 

In  like  manner  J1  will  coincide  with  J. 
Hence  the  prisms  AJ  and  A'J'  coincide  in  all  points. 

.*.  AJ=A'Jf.  Art.  47. 

611.  COR.  1.  Two  truncated  prisms  are  equal  if  tlie 
three  faces  including  a  trihedral  angle  of  one  are  equal  to 
the  three  faces  including  a  trihedral  angle  of  the  other. 


PRISIMS  367 

612.  COR.  2.    Two  right  prisms  are  equal  if  they  have 
equal  l)ases  and  equal  altitudes. 

0    PROPOSITION  IV.     THEOREM 

613.  An  oblique  prism  is  equivalent  to  a  right  prism 
whose  base  is  a  right  section  of  the  oblique  prism  and  whose 
altitude  is  equal  to  a  lateral  edge  of  the  oblique  prism. 


K 


Given  the  oblique  prism  AD',  with  the  right  section  FJ; 
also  the  right  prism  FJf  whose  lateral  edges  are  each  equal 
to  a  lateral  edge  of  AD' . 

To  prove  AD'  =0=  FJ'. 

Proof.  AA  =  FF'.  Hyp. 

Subtracting  FA  from  each  of  these,  AF=A'F'.  (Why?) 

Similarly  BO  =  B'Gr. 

Also  AB  =  AB',  and  FG=F'G'.  Art.  155. 

And  A  of  face  AG  =  homologous  A  of  face  A'O'.  Art.  130. 

/.  face  A6r  =  face  A'G',  Art.  47. 

(for  they  have  all  their  parts  equal,  each  to  each,  and  .".  can  be  made 
to  coincide). 

In  like  manner  face  AK=facQ  A'K' . 
But  face  AD  =  face  AD'.  Art.  591,  3. 

.*.  truncated  prism  AJ=  truncated  prism  AJ'.  Art.  611. 
To  each  of  these  equals  add  the  solid  FD' . 

;.AD'^FJf.  (why?) 

Q.  £.  P. 


368  BOOK  VII.      SOLID    GEOMETRY 

0  PROPOSITION  V.     THEOREM 

614.    The  opposite  lateral  faces  of  a  parallelopiped  are 
equal  and  parallel. 


Given  the  parallelepiped  AH  with  the  base  AC. 
To  prove  AG  =  aud  \\  EH,  and  AJ=  and  ||  EH. 
Proof.  The  base  AC  is  a  ZZ7  .  Art.  598. 

/.  AB=and  \\  EG.  (Why?) 

Also  the  lateral  face  AJ  is  a.  £17  ,  Art.  591,  2. 

.'.  JLF=and  ||  EJ.  .  (Why  ?) 

/.  Z  J3J^=  Z  0^7.  Art.  538. 

And  £~7AG=£JEH.  Art.162. 

Also  plane  AG  \\  plane  J£#.  Art.  538. 

In  like  manner  it  may  be  shown  that  AJ  and  BH  are 
equal  and  parallel. 

Q.  E.  D. 

615.    COR.    Any  two  opposite  -faces  of  a  parallelopiped 
may  be  taken  as  the  bases. 


Ex.  1.    How  many  edges  has  a  parallelopiped  ?   How  many  faces  ? 
How  many  dihedral  angles  ?    How  many  trihedral  angles  ? 

Ex.  2.    Find  the  lateral  area  of  a  prism  whose  lateral  edge  is  10 
and  whose  right  section  is  a  triangle  whose  sides  are  6,  7,  8  in. 

Ex.  3.    Find  the  lateral  area  of  a  right  prism  whose  lateral  edge  is 
16  and  whose  ba§e  is  a  rhombus  with  diagonals  of  6  arid  8  in. 


PRISMS  369 

0   PROPOSITION  VI.    THEOREM 

616.  A  plane  passed  through  two  diagonally  opposite 
edges  of  a  parallelepiped  divides  the  parallelepiped  into  two 
equivalent  triangular  prisms. 


Given  the  parallelepiped  AH  with  a  plane  passed  through 
the  diagonally  opposite  edges  AF  and  CH,  forming  the  tri- 
angular prisms  ABC-G  and  ADC-K. 

To  prove  ABC-Go ADC-K. 

Proof.  Construct  a  plane  J_  to  one  of  the  edges  of  the 
prism  forming  the  right  section  PQRS,  having  the  diagonal 
PR  formed  by  the  intersection  of  the  plane  FHCA. 

Then  PQ  \\  SR,  and  QR  \\  PS.  (Why  ?) 

/.  PQRS  is  a  ZZ7  .  (Why  ?) 

'  .*.  A  PQR=  A  PSR.  (Why  ?) 

But  the  triangular  prism  ABC-G  =c=  a  prism  whose  base 
is  the  right  section  PQR  and  whose  altitude  is  AF.  Art.  613. 

Also  the  triangular  prism  ADC-K  o  a  prism  whose  base 
is  the  right  section  PSR  and  whose  altitude  is  AF.  (Why?) 

But  the  prisms  having  the  equal  bases,  PQR  and  PSR, 
and  the  same  altitude,  AF,  are  equal.  Art.  612. 

/.  ABC-G  =0=  ADC-K.  Ax  l. 

Q.  £.  D. 


370  BOOK  VII.       SOLID     GEOMETRY 

0    PROPOSITION  VII.     THEOREM 

617.    If  two  rectangular  parallelepipeds  have  equal  bases, 
they  are  to  each  other  as  their  altitudes. 


Given  the  rectangular  parallelepipeds  Pf  and  P  having 
equal  bases  and  the  altitudes  A'B'  and  AB. 

To  prove  Pf  :  P=A'B'  :  AB. 

CASE  I.  When  the  altitudes  A'B'  and  AB  are  com- 
mensurable. 

Proof.  Find  a  common  measure  of  A'B'  and  AB,  as 
AK,  and  let  it  be  contained  in  A'B'  n  times  and  in  AB  m 
times. 

Then  A'B'  :  AB  =  n  :  m. 

Through  the  points  of  division  of  A'B'  and  AB  pass 
planes  parallel  to  the  bases. 

These  planes  will  divide  P'  into  n,  and  P  into  m  small 
rectangular  parallelepipeds,  all  equal.  Art.  612. 

/.  P  :  P=n  :  m. 
:.  P'  :  P=A'B'  :  AB.  (Why  ?) 

CASE  II.  When  the  altitudes  A'B'  and  AB  are  incom- 
mensurable. 

Let  the  pupil  supply  the  proof,  using  the  method  of 
limits.  (See  Art.  554). 

618.  DEF.  The  dimensions  of  a  rectangular  parallele- 
piped are  the  three  edges  which  meet  at  one  vertex. 


PAKALLELOPIPEDS 


371 


619.  COR.    If  two  rectangular  parallelopipeds  have  two 
dimensions  in  common,  they  are  to  each  other  as  their  third, 
dimensions. 

PROPOSITION  VIII.     THEOREM 

620.  Two  rectangular  parallelopipeds  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 


Given  the  rectangular  parallelopipeds  P  and  Pf  having 
the  common  altitude  a,  and  the  dimensions  of  their  bases 
b,  c  and  bf,  c',  respectively. 

P      bXc 

To  prove  =  ' 


Then 


Art.  619. 


Proof.  Construct  the  rectangular  parallelepiped,  Q,  whose 
altitude  is  a  and  the  dimensions  of  whose  base  are  b  and  c'. 

P=c_ 
Q    c'' 

AISO  J  =  |7.  (Why?) 

Multiplying  the  corresponding  members  of  these 
equalities, 

£  =  -£*£..  Ax.  4. 

f     b'Xc' 

Q.  E.  D. 

621.  COR.  Two  rectangular  parallelopipeds  having  one 
dimension  in  common  are  to  each  other  as  the  products  of  the 
other  two  dimensions. 


372 


BOOK   VII.       SOLID     GEOMETRY 


PROPOSITION  IX.     THEOREM 

622.    Any  two  rectangular  parallelepipeds  are  to  each 
other  as  the  products  of  their  three  dimensions. 


Given  the  rectangular  parallelepipeds  P  and  Pf  having 

the  dimensions  a,  5,  c  and  af,  ~b' ',  c',  respectively. 

P       aXbXc 


Proof.    Construct  the  rectangular  parallelepiped  Q  hav- 
ing the  dimensions  a,  &,  cf '. 

P=c_ 
Q    c'' 

Q  =  aX  b 
P1     a'  X  I'' 

Multiplying     the    corresponding    members     of     these 
equalities, 

P         n .  V  7)  V  r. 

Ax.  4. 

Q.  E.  B. 


Then 
Also 


Art.  619, 


Art.  621. 


P  ^  a XbX  c 
P'     o'X&'Xc' 


Ex.  1.    Find  the  ratio  of  the  volumes  of  two  rectangular  parallelo- 
pipeds  whose  edges  are  5,  6,  7  in.  and  7,  8,  9  in. 

Ex.2.    Which    will    hold    more,  a    bin    10x2x7   ft.,   or    one   8x 
4  x  5  ft.  ? 

Ex.  8.   How  many  bricks  8  x  4  x  2  in.  are  necessary  to  build  a  wall 
80x6  ft,  x8  in.  f 


PAKALLELOPIPEDS  373 

PROPOSITION  X.     THEOREM 

623.    The  volume  of  a  rectangular  parallelopiped  is  equal 
to  the  product  of  its  three  dimensions. 


u 


Given  the  rectangular  parallelopiped  P  having  the  three 
dimensions  a,  6,  c. 

To  prove          volume  of  P=a  Xb  Xc. 

Proof.    Take  as  the  unit  of  volume  the  cube   Z7,  whose 
edge  is  a  linear  unit. 

P     aX  bXc 

rhen  rTxTxV  Art'622' 

The  volume  of  P  is  the  number  of  times  P  contains  the 

p 
unit  of  volume   U,  or  —  •  Art.  603. 

/.  volume  of  P=a  X  b  X  c. 

Q.  E.  D. 

For  significance  of  this  result,  see  Art.  2. 

624.  COR.  1.    The  volume  of  a  cube  is  the  cube  of  its 
edge. 

625.  COR.  2.    The  volume  of  a  rectangular  parallelo- 
piped is  equal  to  the  product  of  its  base  by  its  altitude. 


Ex.  1.  Find  the  number  of  cubic  inches  in  the  volume  of  a  cube 
whose  edge  is  1  ft.  3  in.  How  many  bushels  does  this  box  contain,  if 
lbushel  =  2150.42cu.  in.  ? 

Ex.  2.  The  measurement  of  the  volume  <  f  a  cube  reduces  to  the 
measurement  of  the  length  of  what  single  straight  line  ? 


374 


BOOK   VII.       SOLID     GEOMETRY 


PROPOSITION  XI.     THEOREM 

626.    The  volume  of  any  parallelopiped  is  equal  to  the 
product  of  its  base  ly  its  altitude. 


Given  the  oblique  parallelopiped  P,  with  its  base  denoted 
by  B,  and  its  altitude  by  H. 

To  prove  volume  of  P=B  X  H. 

Proof.  In  P  produce  the  edge  CD  and  all  the  edges 
parallel  to  CD. 

On  CD  produced  take  FG=CD. 

Pass  planes  through  F  and  G  J_  the  produced  edges, 
forming  the  parallelopiped  Q,  with  the  rectangular  base 
denoted  by  Bf. 

Similarly  produce  the  edge  GI  and  all  the  edges  ||  GI. 

Take  IK=  GI,  and  pass  planes  through  7  and  K  J_  the 
edges  last  produced,  forming  the  rectangular  parallelopiped 
R,  with  its  base  denoted  by  B'f. 


Then 
Also 
But 


Or 

[Outline  Proof. 


volume  of  R  =  B"  X  H. 
volume  of  P=B"XH. 
volume  of  P=BX  H. 


^R=B"  XH=BX  H.] 


Art.  613. 

Art.  386. 

Art.  625. 

Ax.  1. 

Ax.  8. 
Q.  £.  P. 


'PRISMS  375 

PROPOSITION  XII.     THEOREM 


627.    The  volume  of  a  triangular  prism  is  equal  to  the 
product  of  its  base  by  its  altitude. 


Given   the  triangular  prism  PQR-M,  with   its  volume 
denoted  by  F,  area  of  base  by  J5,  and  altitude  by  H. 

To  prove  V=BXH. 

Proof.    Upon  the  edges  PQ,    QR,    QM,   construct   the 
parallelepiped  QK. 

Hence                   QK=  twice  PQR-M.  Art.  616. 

But  volume  of    QK=  area  PQRT  X  H.  Art.  626. 

=  2BXH.  Ax.  8. 

.'.  twice  volume  PQR-M=2B  X  H.  Ax.  i. 

.*.  volume  PQR-M=B  X  H.  Ax.  5. 

Q.  E.  D. 

Ex.  1.    If  the  altitude  of  a  triangular  prism  is  18  in.,  and  the  base 
is  a  right  triangle  whose  legs  are  6  and  8  in.,  find  the  volume. 

Ex.  2     Find  the  volume  of  a  triangular  prism  whose  altitude  is  24, 
and  the  edges  of  whose  base  are  7,  8,  9.     Also  find  the  total  surface. 


tfc     $- -c  • 


376  BOOK  VII.      SOLID    GEOMETRY 

o    PROPOSITION  XIII.     THEOREM 

628.    The  volume  of  any  prism  is  equal  to  the  product  of 
its  base  by  its  altitude. 

L 


Given  the  prism  AK,  with  its  volume  denoted  by   F, 
area  of  base  by  B,  and  altitude  by  H. 

To  prove  V=BXH. 

Proof.    Through  any  lateral  edge,  as  AF,  and  the  diag- 
onals of  the  base,  AC  and  AD,  drawn  from  its  foot,  pass 
'planes. 

These   planes   will    divide    the    prism    into   triangular 
prisms. 

Then  F,  the  volume  of  the  prism  AK,  equals  the  sum 
of  the  volumes  of  the  triangular  prisms.  Ax.  6. 

But  the  volume  of  each  triangular  prism  =  its  base  X  H. 

Hence  the  sum  of  the  volumes  of  the  A  prisms  =  the 
sum  of  the  bases  of  the  A  prisms  X  H. 

=  BXH.  Ax.  8. 

/.  V=BXH.  AX.  l. 

Q.  E.  D. 

629.  COR.  1.    Two  prisms  are  to  each  other  as  the  pro- 
ducts of  their  bases  by  their  altitudes;  prisms  having  equiva- 
lent bases  and  equal  altitudes  are  equivalent. 

630.  COR.  2.    Prisms   having   equivalent   bases   are    to 
each  other  as  their  altitudes;  prisms  having  equal  altitudes 
are  to  each  other  as  their  bases. 


PYKAMIDS 


377 


Pyramid 


0        PYRAMIDS 

631.  A    pyramid    is    a    polyhedron 
bounded  by  a  group  of  planes  passing 
through  a  common  point,  and  by  another 
plane  cutting  all  the  planes  of  the  group. 

632.  The  base  of  a  pyramid  is  the 
face  formed  by  the  cutting  plane;    the  €^~ 
lateral  faces  are  the  faces  formed  by  the 
group  of  planes  passing  through  a  com- 
mon point;   the  vertex  is  the  common  point  through  which 
the  group  of  planes  passes ;   the  lateral  edges  are  the  inter- 
sections of  the  lateral  faces. 

The  altitude  of  a  pyramid  is  the  perpendicular  from  the 
vertex  to  the  plane  of  the  base. 

The  lateral  area  is  the  sum  of  the  areas  of  the  lateral 
faces. 

633.  Properties  of  pyramids  inferred  immediately. 

1.  The  lateral  faces  of  a  pyramid  are  triangles  (Art. 
508,  2). 

2.  The  base  of  a  pyramid  is  a  polygon  (Art.  508,  2). 

634.  A  triangular  pyramid  is  a  pyramid  whose  base  is 
a  triangle;    a  quadrangular  pyramid  is  a  pyramid  whose 
base  is  a  quadrilateral,  etc. 

A  triangular  pyramid  is  also  called  a  tetrahedron,  for  it  has  four 
faces.  All  these  faces  are  triangles,  and  any  one  of  them  may  be 
taken  as  the  base. 


635.  A  regular  pyramid 
is  a  pyramid  whose  base  is  a 
regular  polygon,  and  the  foot 
of  whose  altitude  coincides 
with  the  center  of  the  base. 


378  BOOK   VII.       SOLID     GEOMETRY 

636.  Properties  of  a  regular  pyramid  inferred  immedi- 
ately. 

1.  The  lateral  edges  of  a  regular  pyramid  are  equal,  for 
they  are  oblique  lines  drawn   from   a   point  to   a   plane 
cutting   off   equal    distances   from   the   foot   of    the   per- 
pendicular from  the  point  to  the  plane  (Art.  518). 

2.  The  lateral  faces  of  a  regular  pyramid  are  equal  isos- 
celes triangles. 

637.  The  slant  height  of  a  regular  pyramid  is  the  alti- 
tude of  any  one  of  its  lateral  faces. 

The  axis  of  a  regular  pyramid  is  its  altitude. 

638.  A  truncated  pyramid  is  the  portion  of  a  pyramid 
included  between  the  base  and  a  section  cutting  all  the 
lateral  edges. 

639.  A  frustum  of  a  pyramid  is   the 
part  of  a  pyramid  included   between  the 
base  and  a  plane  parallel  to  the  base. 

The  altitude  of  a  frustum  of  a  pyramid  is 
the  perpendicular  distance  between  the  planes  of  its  bases. 

640.  Properties  of  a  frustum  of  a  pyramid  inferred  im- 
mediately. 

1.  The    lateral  faces   of  a  frustum  of  a   pyramid  are 
trapezoids. 

2.  The  lateral  faces  of  a  frustum  of  a  regular  pyramid 
are  equal  isosceles  trapezoids. 

641.  The  slant  height  of  the  frustum  of  a  regular  pyra- 
mid is  the  altitude  of  one  of  its  lateral  faces. 


Ex.  1.    Show  that  the  foot  of  the  altitude  of  a  regular  pyramid 
coincides  with  the  center  of  the  circle  circumscribed  about  the  base. 

Ex.  2.    The  perimeter  of  the  midsection  of  the  frustum  of  a  pyra- 
mid equals  one -half  the  sum  of  the  perimeters  of  the  bases, 


PYKAMIDS  379 

0    PROPOSITION  XIV.     THEOREM 

642.  The  lateral  area  of  a  regular  pyramid  is  equal  to 
half  the  product  of  the  slant  height  by  the  perimeter  of  the 
base. 

o 


Given  0-ABCDF  a  regular  pyramid  with  its  lateral 
&rea  denoted  by  8,  slant  height  by  L,  and  perimeter  of  its 
base  by  P. 

To  prove  S=%LXP. 

Proof.  The  lateral  faces  GAB,  OBC,  etc.,  are  equal 
isosceles  A.  Art.  636,  2. 

Hence  each  lateral  face  has  the  same  slant  height,  L. 
.'.  the  area  of  each  lateral  face  =  J  L  X  its  base. 
.'.  the  sum  of  all  the  lateral  faces  =  J  L  X  sum  of  bases. 


.  643.  COR.  The  lateral  area  of 
Hie  frustum  of  a  regular  pyramid  is 
equal  to  one-half  the  sum  of  the 
perimeters  of  its  bases  multiplied 
by  its  slant  height. 


Ex.  Find  the  lateral  area  of  a  regular  square  pyramid  whose  slant 
height  is  32,  and  an  edge  of  whose  base  is  16.  Find  the  total 
area  also, 


5 


380 


BOOK   VII.       SOLID     GEOMETRY 


PROPOSITION  XV.     THEOREM 
644.    If  a  pyramid  is  cut  by  a  plane  parallel  to  the  lase, 

I.  The  lateral  edges  and  the  altitude  are  divided  pro- 
portionally; 

II.  The  section  is  a  polygon  similar  to  the  base. 


Given  the  pyramid  S-ABCDF,  with  the  altitude  SO  cut 
by  a  plane  MN,  which  is  parallel  to  the  base  and  intersects 
the  lateral  edges  in  a,  b,  c,  d,  /and  the  altitude  in  o. 

Sa      Sb      Sc  So 

To  prove  I.    —  =— =-^=         =  ^Q' 

II.    The    section    abcdf    similar   to    the    base 
ABCDF. 

Proof.   I.  Pass  a  plane  through  the  vertex  8  II  MN. 
Then  SA,  SB,  SC .  .  .  SO,  are  lines  intersected  by  three 
||  planes. 

•    -^=_^=_^L=  .  .  .  =^L.  Art.  539. 

"  SA     SB     SC  SO 

II.  ab  ||  AB.  (Why?) 

.*.  A  Sab  and  SAB  are  similar.  Art.  328. 

In  like  manner  the  A  Sbc,  Scd,  etc.,  are  similar  to  the 
A  SBC,  SCD,  etc.,  respectively. 

ab  _  /  Sb  \  _  be  _  /  Sc  \  _  cd  _ 


'  AE 


SB 


\ 
) 


BC     \SC        CD 


PVKAMIDS  381 

That  is,  the  homologous  sides  of  abcdf  and  ABCDF  are 
proportional. 

Also       Z.dbc  =  Z  ABC,   £bcd  =  Z  BCD,  etc.      Art.  538. 
.*.  section  obcdf  is  similar  to  the  base  ABCDF.    Art.  321. 

Q.  E.  D. 

645.  COR.  1.  A  section  of  a  pyramid  parallel  to  the 
base  is  to  the  base  as  the  square  of  its  distance  from  the 
vertex  is  to  the  square  of  the  altitude  of  the  pyramid. 

abcdf        ab 

For  (Why  ?> 

ABCDF 


ab      Sa      So         ab2      So 


But  -  =  -  =  -    /.   -  (Why!) 

AB     8A     SO       AB2    SO2 


(Whyf) 


ABCDF 


646.  COR.  2.  //  two  pyramids  having  equal  altitudes 
are  cut  by  a  plane  parallel  to  their  bases  at  equal  distances 
from  the  vertices,  the  sections  have  the  same  ratio  as  the 
bases. 

Let  S-ABCDF  and  V-PQR  be  two  pyramids  cut  as 
described. 


Then  =          (Art.  645);  also  =         .  (          } 

ABCDF    so2  PQR    ~VTL 


But  VT=SO,  and  Vt  =  So.  (Why?) 

pqr          abcdf    ABCDF 
=  °r  -        -  --  (Why  ?) 


647.  COR.  3.  If  two  pyramids  have  equal  altitudes  and 
equivalent  bases,  sections  made  by  planes  parallel  to  the  bases 
at  equal  distances  from  the  vertices  are  equivalent. 


382 


BOOK  VII.      SOLID     GEOMETRY 


PROPOSITION  XVI.     THEOREM 


648.  The  volume  of  a  triangular  pyramid  is  the  limit  of 
the  sum  of  the  volumes  of  a  series  of  inscribed,  or  of  a  series 
of  circumscribed  prisms  of  equal  altitude,  if  the  number  of 
prisms  be  indefinitely  increased. 

r     o 


Given  the  triangular  prism  O-ABC  with  a  series  of  in- 
scribed, and  also  a  series  of  circumscribed  prisms,  formed  by 
passing  planes  which  divide  the  altitude  into  equal  parts,  and 
by  making  the  sections  so  formed  first  upper  bases,  then 
lower  bases,  of  prisms  limited  by  the  next  parallel  plane. 

To  prove  O-ABC  the  limit  of  the  sum  of  each  series,  if 
the  number  of  prisms  in  each  be  indefinitely  increased. 

Proof.  Each  inscribed  prism  equals  the  circumscribed 
prism  immediately  above  it.  Art.  629. 

.*.  (sum  of  circumscribed  prisms) — (sum  of  inscribed 
prisms)  =  lowest  circumscribed  prism,  or  ABC-K. 

If  the  number  of  prisms  be  indefinitely  increased,  the 
altitude  of  each  approaches  zero  as  a  limit. 

Hence  volume  ABC-K=0,  Art.  253,  2. 

(for  its  base,  ABC,  is  constant  while  its  altitude  =  0). 

.*.  (sum  of  circumscribed  prisms) — (sum  of  inscribed 
prisms)  =  0. 

.*.  volume  O-ABC — (either  series  of  prisms)  =  0. 

(for  this  difference  <  difference  between  the  two  series,  which 

last  difference  =0). 

/.  O-ABC  is  the  limit  of  the  sum  of  the  volumes  of 
either  series  of  prisms.  Q.  E.  D. 


PYRAMIDS 


PROPOSITION  XVII.      THEOREM 


383 


649.    If  two   triangular  pyramids  have  equal  altitudes 
and  equivalent  bases,  they  are  equivalent. 

o 


B 


Given  the  triangular  pyramids  O-ABC  and  O'-A'B'C' 
having  equivalent  bases  ABC  and  A'B'O ',  and  equal 
altitudes. 

To  prove  O-ABC-  O'-A'B'C'. 

Proof.  Place  the  pyramids  so  that  they  have  the  com- 
mon altitude  H,  and  divide  H  into  any  convenient  num- 
ber of  equal  parts. 

Through  the  points  of  division  and  parallel  to  the  plane  of 
the  bases  of  the  pyramids,  pass  planes  cutting  the  pyramids. 

Using  the  sections  so  formed  as  upper  bases,  inscribe  a 
series  of  prisms  in  each  pyramid,  and  denote  the  volumes 
of  the  two  series  of  prisms  by  V  and  V. 

The  sections  formed  by  each  plane,  as  KLM  and  K'L'M', 
are  equivalent.  Art.  647. 

.".   each  prism  in    O-ABC  =e=  corresponding  prism  in 

Of-A'B'Cf  (Art.  629).  /.    V=  Vf.  Ax.  2. 

Let  the  number  of  parts  into  which  the  altitude  is 
divided  be  increased  indefinitely. 

Then  V  and  V  become  variables  with  O-ABC  and 
O'-A'B'C'  as  their  respective  limits.  Art.  648. 

But  V^V  always.  (Why?) 

.-.  O-ABC^O'-A'B'C',  (Why?) 

Q.  E.  D. 


384 


BOOK   VII.      SOLID     GEOMETRY 


PROPOSITION  XVIII.     THEOREM 

650.    The  volume  of  a  triangular  pyramid  is  equal  tc 
one-third  the  product  of  its  base  ly  its  altitude. 


D 


Given  the  triangular  pyramid  0-ABC,  having  its  volume 
denoted  by  V,  the  area  of  its  base  by  B,  and  its  altitude  by  27". 

To  prove  V=$BXH. 

Proof.  On  ABC  as  a  base,  with  OB  as  a  lateral  edge, 
construct  the  prism  ABC-DOF. 

Then  this  prism  will  be  composed  of  the  original  pyra- 
mid O-ABC  and  the  quadrangular  pyramid  O-ADFC. 

Through  the  edges  OD  and  OC  pass  a  plane  intersecting 
the  face  ADFC  in  the  line  DC,  and  dividing  the  quadrangular 
pyramid  into  the  triangular  pyramids  0-ADC  and  0-DFC. 

Then  0-ADC^  0-DFC.  Art.  649. 

(for  they  have  the  common  vertex  O,  and  the  equal  bases  ADC  and  DFC}. 

But  0-DFC  may  be  regarded  as  having  C  as  its  vertex 
and  DOF  as  its  base.  Art.  634. 

/.  0-DFC^O-ABC.  Art.  649.* 

.*.  the  prism  is  made  up  of  three  equivalent  pyramids. 

.*.  0-ABC= J  the  prism.  ^  Ax.  5. 

But  volume  of  prism  =  B  X  H.  Art.  627. 

.-.  0-ABC,  or  F=J  B  X  H.  Ax.  5. 

0.  E.  D. 

Ex.  Find  the  volume  of  a  triangular  pyramid  whose  altitude  is  12 
ft.,  and  whose  base  is  an  equilateral  triangle  with  a  side  of  15  ft. 


PYRAMIDS  385 


PROPOSITION  XIX.     THEOREM 

651.    The  volume  of  any  pyramid  is  equal  to  one-third 
the  product  of  its  base  by  its  altitude. 


Given  the  pyramid  0-ABCDF,  having  its  volume  denoted 
by  V,  the  area  of  its  base  by  B,  and  its  altitude  by  H. 

To  prove  V=$BXH. 

Proof.  Through  any  lateral  edge,  as  OD,  and  the  diago- 
nals of  the  base  drawn  from  its  foot,  as  AD  and  BD,  pass 
planes  dividing  the  pyramid  into  triangular  pyramids. 

Then  V,  the  volume  of  the  pyramid  0-ABCDF,  will 
equal  the  sura  of  the  volumes  of  the  triangular  pyramids. 

But  the  volume  of  each  A  pyramid  =  &  its  base  X  H. 

Art.  650. 

Hence  the  sum  of  the  volumes  of  A  pyramids  =  J  sum  of 

their  bases  X  H.  Ax.  2. 

=  %BXH.  Ax.  8. 

.-.  V=$  BX  H.  Ax.  1. 

Q.  E.  D. 

652.  COR.  1.    The  volumes  of  two  pyramids  are  to  each 
other  as  the  products  of  their  bases  and  altitudes;   pyramids 
having  equivalent  bases  and  equal  altitudes  are  equivalent. 

653.  COR.  2.    Pyramids  having  equivalent  bases  are  to 
each  other  as  their  altitudes;   pyramids  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 

654.  SCHOLIUM.    The  volume  of  any  polyhedron  may  be 
found  by  dividing  the  polyhedron  into  pyramids,  finding  the 
volume  of  each  pyramid  separately,  and  taking  their  sum. 


386 


BOOK   VII.       SOLID     GEOMETRY 


0      PROPOSITION  XX.     THEOREM 

655.  The  frustum  of  a  triangular  pyramid  is  equivalent 
to  the  sum  of  three  pyramids  whose  common  altitude  is  the 
altitude  of  the  frustum,  and  whose  bases  are  the  lower  base, 
the  upper  base,  and  a  mean  proportional  between  the  two 
bases  of  the  frustum . 


Given  ABC-DEF  the  frustum  of  a  triangular  pyramid, 
having  the  area  of  its  lower  base  denoted  by  B,  the  area 
of  its  upper  base  by  b,  and  its  altitude  by  H. 

To  prove  ABC-DEF  =c=  three  pyramids  whose  bases  are 
B,  b  and  vBb,  and  whose  common  altitude  is  H. 

Proof.  Through  E  and  A  C,  E  and  DC,  pass  planes  divid- 
ing the  frustum  into  three  triangular  pyramids.  Then 

1.  E-ABC  has  the  base  B  and  the  altitude  H. 

2.  E-DFC,  that  is,   C-DEF,  has  the  base  b  and   the 
altitude  H. 

3.  It  remains  to  show  that  E-ADC  is  equivalent  to  a 
pyramid  having  an  altitude  H,  and  a  base  that  is  a  mean 
proportional  between  B  and  b. 

Denoting  the  three  pyramids  by  I,  II,  III, 

I=A  ABE=AB  =  AC=&ADC=H 

II     A  ADE    DE    DF    A  DFC    III" 
(Arts.  653,  391,  644,  321.     Let  the  pupil  supply  the  reason  for  each 


I       II 


step  in  detail). 


or 


=VlXIII.  Art.  303. 


b. 
Hence,  ABC-DEFosum  of  three  pyramids,  as  described. 

Q.  £.  D. 


PYRAMIDS 


387 


656.    Formula  for  volume  of  frustum  of   a  triangular 
pyramid.  y=  J H  (B  +  b  + 


PROPOSITION  XXI.     THEOREM 

657.  The  volume  of  the  frustum  of  any  pyramid  is 
equivalent  to  the  sum  of  the  volumes  of  three  pyramids,  whose 
common  altitude  is  the  altitude  of  the  frustum,  and  ivhose 
bases  are  the  lower  base,  the  upper  base,  and  a  mean  propor- 
tional between  the  two  bases  <f  the  frustum. 

K  T 

A 

T\ 

^ 


yt 


Given  the  frustum  of  a  pyramid  Ad,  having  its  volume 
denoted  by  V,  the  area  of  its  lower  base  by  B,  of  its  upper 
base  by  5,  and  its  altitude  by  H. 

To  prove         V=  J  H  (B  +  b  - 

Proof.    Produce  the  lateral  faces  of  Ad  to  meet  in  K. 

Also  construct  a  triangular  pyramid  with  base  PQR 
equivalent  to  ABCDF,  and  in  the  same  plane  with  it,  and 
with  an  altitude  equal  to  the  altitude  of  K-ABCDF.  Pro- 
duce the  plane  of  ad  to  cut  the  second  pyramid  in  pqr. 

Then  pqr^abcdf.  Art.  647.  \ 

:.  pyramid  K-ABCDF  ^  pyramid  T-PQR.     Art.  652. 

Also        pyramid  K-abcdf  o  pyramid  T-pqr.  (Why  ?) 

Subtracting,  frustum  Ad  =c=  frustum  Pr.  Ax.  3. 

But  volume  Pr  =  $  H  (B  +  b  +  V  Eb) . 

;.  volume  Ad  =  J  H  (B  +  b  +  V '  Bb) .     (Why  ?) 

Q.  £.  D. 


388  BOOK  VII.       SOLID     GEOMETRY 

PROPOSITION  XXII.     THEOREM 

658.  A  truncated  triangular  prism  is  equivalent  to  the 
sum  of  three  pyramids,  of  which  the  base  of  the  prism  is  the 
common  base,  and  ivhose  vertices  are  the  three  vertices  of  the 
inclined  section. 


,R 


Fig.  1  Fig.  2 


Given  the  truncated  triangular  prism  ABC-PQR, 

To  prove  ABC-PQR  =  the  sura  of  the  three  pyramids 
P-ABC,  Q-ABC  and  R-ABC. 

Proof.  Pass  planes  through  Q  and  AC,  Q  and  PC,  divid- 
ing the  given  figure  into  the  three  pyramids  Q-ABC, 
Q-APC  and  Q-PRC. 

1.  Q-ABC  has  the    required   base    and   the   required 
vertex  Q. 

2.  Q-APC  =0  B-APC,  Art.  652. 
(for  they  have  the  same  base,  A  PC,  and  the  same  altitude,  their  ver- 
tices being  in  a  line  \\  base  APC). 

But  B-APC  may  be  regarded  as  having  P  for  its  ver- 
tex, and  ABC  for  its  base,  as  desired.  Art.  634. 

3.  Q-PRC  =0  B-ARC  (see  Fig.  2) .  Art.  652. 

{for  the  base  ARC  ^  base  PRC  (Art.  :WO);  and  the  altitudes  of  the  two 

pyramids  are  equal,  the  vertices  Q  ami  />'  hcimj  in  line  \\  plane 

PACE,  in  'which  the  bases  tie). 

But  B-ARC  may  be  regarded  as  having  R  for  its  ver- 
tex, and  ABO  for  its  base,  as  desired.  Art,  634, 


PKISMATOIDS 


389 


.'.  ABC-PQR  =0=  sum  of  three  pyramids  whose  common 
base  is  ABC,  and  whose  vertices  are  P,  Q,  JR. 

Q.  E.  D. 


Pig.  3 


Fig.  4 


659.  COR.  1.    The  volume  of  a  truncated  right  triangu- 
lar prism  (Fig.  3)  is  equal  to  the  product  of  its  base  by 
one- third  the  sum  of  its  lateral  edges. 

660.  COR.  2.    The  volume  of  any  truncated  triangular 
prism   (Fig.  4)  is  equal  to  the  product  of  the  area  of  its 
right  section  by  one- third  the  sum  of  its  lateral  edges. 


PRISMATOIDS 

661.  A    prismatoid    is     a    polyhedron 
bounded  by  two  polygons  in  parallel  planes, 
called  bases,  and  by  lateral  faces  which  are 
either  triangles,  trapezoids  or  parallelograms. 

662.  A   prismoid    is   a    prismatoid    in 
which  the  bases  have  the  same  number  of 
sides  and   have  their  corresponding   sides 
parallel. 


Erlamoid 


Ex.  The  volume  of  a  truncated  right  parallelepiped  equals  the 
area  of  the  lower  base  multiplied  by  oue- fourth  the  sum  of  the  lateral 
edges  (or  by  a  perpendicular  from  the  center  of  the  upper  base  to  tiie 
Ipwer  base), 


390 


BOOK  VII.       SOLID     GEOMETKY 


PROPOSITION  XXIII.     THEOREM 

663.  The  volume  of  a  prismatoid  is  equal  to  one-sixth 
the  product  of  its  altitude  ~by  the  sum  of  its  bases  and  of 
four  times  the  area  of  its  midsection. 


G 


Given  the  prismatoid  ABCD-FGK,  with  bases  B  and  b, 
midsection  M,  volume  F,  and  altitude  H. 

To  prove  F=i  H  (B  +  I  +  4  M). 

Proof.  Take  any  point  0  in  the  midsection,  and  through 
it  and  each  edge  of  the  prismatoid  let  planes  be  passed. 

These  planes  will  divide  the  figure  into  parts  as  follows: 

1.  A  pyramid  with  vertex  0,  base  ABCD  and  altitude 
J  _5T,  and  whose  volume  /.  =i  SX  B.  Art.  651. 

2.  A  pyramid  with  vertex  0,  base  FGK,  and  altitude  J 
H,  and  whose  volume  .*.  =i  J3"X  &.  (Why?) 

3.  Tetrahedrons   like  0-ABG  whose  volume  may  be 
determined  as  follows  (see  Fig.  2)  : 

AB  =  2  PQ.  (Why?) 

.-.  A  AGB  =  ±  A  PGQ.  Art.  398. 

/.   0-AGB  =  4  0-PGQ.  Art.  653. 

But  0-PGQ  (or  G-PQO)  =  i  PQO  X  J  fi=i  HXPQO. 

:.  0-AGB  =  i  HX4A  PQO.        (Why  ?) 

.'.  the  sum  of  all  tetrahedrons  like  0-A  GB  =  i  jff  X  4  M. 


Or 


KEGULAK     POLYHEDKONS  391 

REGULAR    POLYHEDRONS 

664.  DEF.    A  regular  polyhedron  is  a  polyhedron  all  of 
whose  faces  are  equal  regular  polygons,  and  all  of  whose 
polyhedral  angles  are  equal.     Thus,  the  cube  is  a  regular 
polyhedron. 

PROPOSITION  XXIV.     THEOREM 

665.  But  five  regular  polyhedrons  are  possible. 


Given  regular  polygons  of  3,  4,  5,  etc.,  sides. 

To  prove  that  regular  polygons  of  the  same  number  of 
sides  can  be  joined  to  form  polyhedral  A  of  a  regular 
polyhedron  in  but  five  different  ways,  and  that,  conse- 
quently, but  five  regular  polyhedrons  are  possible. 

Proof.  The  sum  of  the  face  A  of  any  polyhedral  angle 
<  360°.  Art.  583. 

1.  Each  Z  of  an  equilateral  triangle  is  60°.         Art.  134. 
3  X  60°,  4  X  60°  and  5  X  60°  are  each  less  than  360°; 

but  any  larger  multiple  of  60°  =  or  >  360°. 

.*.  but  three  regular  polyhedrons  can  be  formed  with 
equilateral  A  as  faces. 

2.  Each  Z  of  a  square  contains  90°.  Art.  151. 
3  X  90°  is  less  than  360°,  but  any  larger  multiple  of 

90°  =  or  >  360°. 

.*.  but  one  regular  polyhedron  can  be  formed  with 
squares  as  faces. 

3.  Each  Z  of  a  regular  pentagon  is  108°.  Art.  174. 
3  X  108°  is  less  than  360°,  but  any  larger  multiple  of 

108°  >  360°. 


392 


BOOK   VII.       SOLID     GEOMETRY 


/.  but  one  regular  polyhedron  can  be  formed  with  regu* 
lar  pentagons  as  faces. 

4.  Each  Z  of  a  regular  hexagon  is  120°,  and  3  X  120° 
-360°. 

.*.  no  regular  polyhedron  can  be  formed  with  hexagons, 
or  with  polygons  with  a  greater  number  of  sides  as  faces. 

/.  but  five  regular  polyhedrons  are  possible. 

Q.  E.  D. 

666.  The  construction  of  the  regular  polyhedrons,  by  the 
use  of  cardboard,  may  be  effected  as  follows: 

Draw  on  a  piece  of  cardboard  the  diagrams  given  below. 
Cut  the  cardboard  half  through  at  the  dotted  lines  and  en- 
tirely through  at  the  full  lines.  Bring  the  free  edges 
together  and  keep  them  in  their  respective  positions  by 
some  means,  such  as  pasting  strips  of  paper  over  them. 


Octahedron 


Tetrahedron 


Hexahedron 


Dodecahedron 


Icosahedron 


POLYHEDRONS 


393 


POLYHEDRONS    IN    GENERAL 

PROPOSITION  XXV.     THEOREM 

667.  In  any  polyhedron,  the  number  of  edges  increased 
by  two  equals  the  number  of  vertices  increased  by  the  number 
of  faces.  p 


Given  the  polyhedron  AT,  with  the  number  of  its  ver- 
tices, edges  and  faces  denoted  by  F,  E  and  F,  respectively. 

To  prove  E  +  2=V+F. 

Proof.  Taking  the  single  face  ABCD,  the  number  of 
edges  equals  the  number  of  vertices,  or  E=  V. 

If  another  face,  CRTD,  be  annexed  (Fig.  2),  three  new 
edges,  CR,  RT,  TD,  are  added  and  two  new  vertices,  R  and  T. 

:.  the  number  of  edges  gains  one  on  the  number  of  ver- 
tices, or  E=V+  1. 

If  still  another  face,  BQRC,  be  annexed,  two  new  edges, 
BQ  and  QR,  are  added,  and  one  new  vertex,  Q.  .*.  E=  F+2. 

With  each  new  face  that  is  annexed,  the  number  of 
edges  gains  one  on  the  number  of  vertices,  till  but  one 
face  is  lacking. 

The  last  face  increases  neither  the  number  of  edges  nor 
of  vertices. 

Hence  number  of  edges  gains  one  on  number  of  vertices, 
for  every  face  except  two,  the  first  and  the  last,  or  gains 
F—2  in  all. 

.*.  for  the  entire  figure,  E=  F  +  F—  2. 

That  is  E  +  2=  V  +  F.  Ax.  2. 

Q.  B.  D. 


394  BOOK  VII.       SOLID     GEOMETRY 

PROPOSITION  XXVI.     THEOREM 

668.  The  sum  of  the  face  angles  of  any  polyhedron 
equals  four  right  angles  taken  as  many  times,  less  two,  as 
the  polyhedron  has  vertices. 


Given  any  polyhedron,  with  the  sum  of  its  face  angles 
denoted  by  S,  and  the  number  of  its  vertices,  edges  and 
faces  denoted  by  F,  E,  F,  respectively. 

To  prove  S=(V—  2)  4  rt.  A. 

Proof.  Each  edge  of  the  polyhedron  is  the  intersection 
of  two  faces,  .'.  the  number  of  sides  of  the  faces  =  2  E.  ^ 

:.  the  sum  of  the  interior  and  exterior  A  of  the  faces  = 
2  E  X  2  rt.  A  ,  or  E  X  4  rt.  A  .  Art.  73. 

But  the  sum  of  the  exterior  A  of  each  face  =  4  rt.  A  . 

Art.  175  „ 

/.  the  sum  of  exterior  A  of  the  FfsLces  =  FX  4  rt.  A  . 

Ax.  4. 

Subtracting  the  sum  of  the  exterior  A  from  the  sum  of 
all  the  A  ,  the  sum  of  the  interior  A  of  the  F  faces  = 
(EXlrt.  A)—(FX4rt.  A). 

Or  S=(E—F)  4rt.  A. 

But                           E  +  2=V+F.  Art.  667. 

Hence                       E—F=  F— 2.  Ax.  3. 

Substituting  for  E—  F,  S=  (  F— 2)  4  rt.  A  .  Ax.  8. 

Q.  E.  D. 

Ex.    Verify  the  last  two  theorems  in  the  case  of  the  cube. 


COMPARISON     OF     POLYHEDRONS 


395 


COMPARISON  OF  POLYHEDRONS.     SIMILAR  POLYHEDRONS 

PROPOSITION  XXVII.     THEOREM 

669.  //  two  tetrahedrons  have  a  trihedral  angle  of  one 
equal  to  a  trihedral  angle  of  the  other,  they  are  to  each  other 
as  the  products  of  the  edges  including  the  equal  trihedral 

angles. 

c 

c' 


Given  the  tetrahedrons  O-ABC  and  0'-A'B'Cr,  with 
their  volumes  denoted  by  Fand  V ',  respectively,  and  having 
the  trihedral  A  0  and  Of  equal. 

V_=     OAXOBX  PC 
V     0'A'XO'B'Y.O'C'' 

Proof.  Apply  the  tetrahedron  O'-A'B'C'  to  O-ABC  so 
that  the  trihedral  Z  0'  shall  coincide  with  its  equal,  the 
trihedral  Z  0. 

Draw  CP  and  G'P'  JL  plane  QAB,  and  draw  OP  the 
projection  of  OC  in  the  plane  OAB. 

Taking  OAB  and  OA'B'  as  the  bases,  and  CP  and  C'P' 
as  the  altitudes  of  the  pyramids  C-OAB  and  C'-OA'B', 
respectively. 


V       A  OAB  X  OP 

&OAB  ^ 

CP 

Art.  652. 
Art.  397. 
(Why  ?) 

But 

In  the 

V     A  OA 
A 

'B'  X  O'P' 
OA£       OA 

X  OB 

C'P' 

OC 

~00'* 
X  00 

A 
similar  rt.  / 
OA  X  6 

OA'B'     OA'  X  0£' 

^7P 

&  OOPandOO'P',  7^7  = 

0  X 

U3  X  00         OA  X  05 

"7 

'     OA'  X  0 

£'  X  00X     C 

VA>  X  O'B' 

X  0'( 

^  .  •     AX  .  o  . 
Q.  £.  D. 

396 


BOOK  vii.     SOLID   GEOMETRY 


670.  DEP.    Similar  polyhedrons  are  polyhedrons  having: 
the  same  number  of  faces,  similar,  each  to  each,  and  simi- 
larly placed,   and  having  their  corresponding  polyhedral 
angles  equal. 

PROPOSITION  XXVIII.     THEOREM 

671.  Any  two  similar  polyhedrons  may  be  decomposed 
into  the  same  number  of  tetrahedrons,  similar,  each  to  each, 
and  similarly  placed. 


c 


Given  P  and  Pf,  two  similar  polyhedrons. 
To  prove  that  P  and  Pf  may   be   decomposed   into   the 
same  number  of  tetrahedrons,  similar,  each  to  each. 

Proof.  Take  H  and  Hf  any  two  homologous  vertices  of 
P  and  P'.  Draw  homologous  diagonals  in  all  the  faces  of 
P  and  P'  except  those  faces  which  meet  at  H  and  Hf,  sepa- 
rating the  faces  into  corresponding  similar  triangles. 

Through  H  and  each  face  diagonal  thus  formed  in  P, 
and  through  R'  and  each  face  diagonal  in  P',  pass  planes. 

Each  corresponding  pair  of  tetrahedrons  thus  formed 
may  be  proved  similar. 

Thus,  in  H-ABC  and  R'-A'B'Cf,  the  A  HBA  and 
R'E'A!  are  similar.  Art.  329. 

In  like  manner  A  REG  and  R'E'G'  are  similar;  and  A 
ABC  and  A'B'C'  are  similar. 


SIMILAR    POLYGONS  397 

AC 

'    Art'321' 


/.  A  ARC  and  A'H'Cf  are  similar.  Art.  326. 

Hence  the  corresponding  faces  of  H-  ABC  and  Hf-A'BfCf 
are  similar. 

Also  their  homologous  trihedral  A  are  equal.      Art.  584. 

•.  tetrahedron  H-ABC  is  similar  to  H'-AfB'Cf.  Art.  670. 

After  removing  H-ABC  from  P,  and  H'-A'B'C'  from 

P',  the  remaining  polyhedrons  are  similar,  for  their  faces 

are  similar,  and  the  remaining  polyhedral  A  are  equal. 

Ax.  3. 

By  continuing  this  process,  P  and  Pf  may  be  decom- 
posed into  the  same  number  of  tetrahedrons,  similar,  each 
to  each,  and  similarly  placed. 

Q.  £.  D. 

672.  COR.  1.    The  homologous  edges  of  similar  polyhe- 
drons are  proportional; 

Any  two  homologous  lines  in  two  similar  polyhedrons 
have  the  same  ratio  as  any  other  two  homologous  lines. 

673.  COB.  2.    Any  two  homologous  faces  of  two  similar 
polyhedrons  are   to  each  other  as  the  squares  of  any  two 
homologous  edges  or  lines; 

The  total  areas  of  any  two  similar  polyhedrons  are  to 
each  other  as  the  squares  of  any  two  homologous  edges. 


Ex.  1.  In  the  figure,  p.  395,  if  the  edges  meeting  at  OareS,  9,  12  in., 
and  those  meeting  at  0'  are  4,  6,  8  in.,  find  the  ratio  of  the  volumes 
of  the  tetrahedrons. 

Ex.  2.  If  the  linear  dimensions  of  one  room  are  twice  as  great  as 
the  corresponding  dimensions  of  another  room,  how  will  their  surfaces 
(and  .'.  cost  of  papering)  compare  ?  How  will  their  volumes  compare  ? 

Ex.  3.    How  many  2  in.  cubes  can  be  cut  from  a  10  in.  cube  ? 

Ex.  4.  If  the  bases  of  a  prismoid  are  rectangles  whose  dimensions 
are  a,  b  and  b,  a,  and  altitude  is  //,  find  the  formula  for  the  volume. 


Vnfjy 


398  BOOK  VII.      SOLID     GEOMETRY 


PROPOSITION  XXIX.     THEOREM 

674.    The  volumes  of  two  similar  tetrahedrons  are  to 
each  other  as  the  cubes  of  any  pair  of  homologous  edges. 


Given  the  similar  tetrahedrons  0-ABC  and  O'-A'B'C'. 


V      OA 

To  prove  —  ^= 

F 


_      ,  V        OA  X  OB  X  OC 

Proof.  -  =  —————.  (Why!) 

OA  x  OS  ^  00 


O'B'     0>0 
OA       OB      OG 

But  =='  Art'672' 


y       0.4  ^  OA  ^  OA     ~OA3 

—  =  -  X  -  X  -  =  Ax.  8. 

F 


Ex.  1.  In  the  above  figures,  if  AB=2  A'B',  find  the  ratio  of  V 
to  F'.  Find  the  same,  if  AB=H  A'B'. 

Ex.  2  The  measurement  of  the  volume  of  a  regular  triangular 
prism  reduces  to  the  measurement  of  the  lengths  of  how  many  straight 
lines  ?  of  a  frustum  of  a  regular  square  pyramid  ? 

Ex.  3.  Show  how  to  construct  out  of  pasteboard  a  regular  prism, 
a  parallelepiped,  and  a  truncated  square  prism. 


SIMILAK    POLYGONS  399 

PROPOSITION  XXX.     THEOREM 

675.  The  volumes  of  any  two  similar  polyhedrons  are 
to  'each  other  as  the  cubes  of  any  two  homologous  edges,  or 
of  any  other  two  homologous  lines. 


BO  B1     O1 

Given  the  polyhedrons  AK  and  A'K'  having  their  vol- 
umes denoted  by  V  and  Vft  and  HB  and  H'B'  any  pair  of 
homologous  edges. 

V  _HB3 
To  prove  —  — 

y    H'B'* 

Proof.  Let  the  polyhedrons  be  decomposed  into  tetra- 
hedrons, similar,  each  to  each,  and  similarly  placed.  Art.  671. 

Denote  the  volumes  of  the  tetrahedrons  in  P  by  vi,  V2, 
vs  .  .  .  and  of  those  in  Pf  by  v'i,  1/2,  v's  •  •  • 


Then 
Also 


HfBfi 


!_!_ 

1/1 


Art.  674,  Ax.  1. 


(for  each  of  these  ratios^. 


Vl 


j±at=:-2Lj  that  is>  *1=£L.    Art.  312. 


HB 


Ax.  1. 
Q.  E.  D- 


400  BOOK  VII.      SOLID     GEOMETRY 

EXERCISES.    CROUP  67 

THEOREMS    CONCERNING    POLYHEDRONS 
Ex.  1.    The  lateral  faces  of  a  right  prism  are  rectangles. 

Ex.  2.  A  diagonal  plane  of  a  prism  is  parallel  to  every  lateral  edge 
of  the  prism  not  contained  in  the  plane. 

Ex.  3.    The  diagonals  of  a  parallelepiped  bisect  each  other. 

Ex.  4.  The  square  of  a  diagonal  of  a  rectangular  parallelepiped 
equals  the  sum  of  the  squares  of  the  three  edges  meeting  at  a  vertex. 

Ex.  5.  Each  lateral  face  of  a  prism  is  parallel  to  every  lateral  edge 
not  contained  in  the  face. 

Ex.  6.  Every  section  of  a  prism  made  by  a  plane  parallel  to  a 
lateral  edge  is  a  parallelogram. 

Ex.  7.  If  any  two  diagonal  planes  of  a  prism  which  are  not  par- 
allel to  each  other  are.  perpendicular  to  the  base  of  the  prism,  the 
prism  is  a  right  prism. 

Ex.  8.  What  part  of  the  volume  of  a  cube  is  the  pyramid  whose 
"base  is  a  face  of  the  cube  and  whose  vertex  is  the  center  of  the  cube  ? 

Ex.  9.  Any  section  of  a  regular  square  pyramid  made  by  a  plane 
through  the  axis  is  an  isosceles  triangle. 

Ex.  10.  In  any  regular  tetrahedron,  an  altitude  equals  three  times 
the  perpendicular  from  its  foot  to  any  face. 

Ex.  11.  In  any  regular  tetrahedron,  an  altitude  equals  the  sum  of 
the  perpendiculars  to  the  faces  from  any  point  within  the  tetrahedron. 

Ex.  12.  Find  the  simplest  formula  for  the  lateral  area  of  a  trun- 
cated regular  prism  of  n  sides. 

Ex.  13.    The  sum  of  the  squares  of  the  four  diagonals  of  a  paral- 
lelopiped  is  equal  to  the  sum  of  the  squares  of  the  twelve  edges. 
[Sue.    Use  Art.  352.] 

Ex.  14.     A  parallelepiped  is  symmetrical  with  respect  to  what  point? 

Ex.  15.  A  rectanglar  parallelepiped  is  symmetrical  with  respect  to 
how  many  planes  ?  (Let  the  pupil  make  a  definition  of  a  figure  sym 
metrical  with  respect  to  a  plane.  See  Arts.  486,  487.) 


EXERCISES  ON  POLYHEDRONS         401 

Ex.  16.  The  volume  of  a  pyramid  whose  lateral  edges  are  the 
three  edges  of  the  parallelepiped  meeting  at  a  point  is  what  part  of 
the  volume  of  the  parallelepiped  ? 

Ex.  17.  If  a  plane  be  passed  through  a  vertex  of  a  cube  and  the 
diagonal  of  a  face  not  adjacent  to  the  vertex,  what  part  of  the  volume 
of  the  cube  is  contained  by  the  pyramid  so  formed  ? 

Ex.  18.  If  the  angles  at  the  vertex  of  a  triangular  pyramid  are 
right  angles  and  each  lateral  edge  equals  a,  show  that  the  volume  of 

the  pyramid  is  — * 
b 

Ex.  19.  How  large  is  a  dihedral  angle  at  the  base  of  a  regular 
pyramid,  if  the  apothem  of  the  base  equals  the  altitude  of  the  pyramid  ? 

Ex.  20.  The  area  of  the  base  of  a  pyramid  is  less  than  the  area  of 
the  lateral  surface. 

Ex.  21.  The  section  of  a  triangular  pyramid  by  a  plane  parallel 
to  two  opposite  edges  is  a  parallelogram. 

If  the  pyramid  is  regular,  what  kind  of  a  parallelogram  does  the 
section  become  ? 

Ex.  22.  The  altitude  of  a  regular  tetrahedron  divides  an  altitude 
of  the  base  into  segments  which  are  as  2  :  1. 

Ex.  23.  If  the  edge  of  a  regular  tetrahedron  is  a,  show  that  the 
slant  height  is  -~-  ;  and  hence  that  the  altitude  is  ^— ,  and  the  vol- 

2t  o 


Ex.  24.  If  the  midpoints  of  all  the  edges  of 
a  tetrahedron  except  two  opposite  edges  be 
joined,  a  parallelogram  is  formed. 

Ex.  25.  Straight  lines  joining  the  midpoints 
of  the  opposite  edges  of  a  tetrahedron  meet  in 
a  point  and  bisect  each  other. 

Ex.  26.    The  midpoints  of  the  edges  of  a  regular  tetrahedron  are, 
the  vertices  of  a  regular  octahedron. 


402  BOOK   VII.      SOLID     GEOMETRY 

EXERCISES.    GROUP  68 

PROBLEMS    CONCERNING    POLYHEDRONS 

Ex.  1 .    Bisect  the  volume  of  a  given  prism  by  a  plane  parallel  to 
the  base. 

Ex.  2.     Bisect  the  lateral  surface  of  a  given  pyramid  by  a  plane  par- 
allel to  the  base. 

Ex.  3.    Through  a  given  point  pass  a  plane  which  shall  bisect  the 
volume  of  a  given  parallelepiped. 

Ex.  4.    Given  an  edge,  construct  a  regular  tetrahedron. 
Ex.  5.    Given  an  edge,  construct  a  regular  octahedron. 

Ex.  6.    Pass  a  plane  through  the  axis  of  a  regular  tetrahedron  so 
that  the  section  shall  be  an  isosceles  triangle. 

Ex.  7.    Pass  a  plane  through  a  cube  so  that  the  section  shall  be  a 
regular  hexagon. 

Ex.  8.    Through  three  given  lines  no  two  of  which  are  parallel 
pass  planes  which  shall  form  a  parallelepiped. 

Ex.  9.    From  cardboard  construct  a  regular  square  pyramid  each 
of  whose  faces  is  an  equilateral  triangle. 


EXERCISES.     CROUP  69 

REVIEW    EXERCISES 

Make  a  list  of  the  properties  of 

Ex.  1.    Straight  lines  in  space.  Ex.  9.    Right  prisms. 

Ex.  2.    One  line  and  one  plane.          Ex.  10.    Parallelopipeds  in  gen- 
Ex.  3.    Two  or   more  lines   and 

one  plane.  Ex.  11.    Rectangular    parallelo- 

Ex.  4.  Two  planes  and  one  line.  pipeds. 

Ex.5.  Two  planes  and  two  lines.  Ex«  12«    Pyramids  in  general. 

Ex.  6.  Polyhedrons  in  general.  Ex-  13-    Regular  pyramids. 

Ex.  7.  Similar  polyhedrons.  Ex.  14     Frusta  of  pyramids. 

Ex.8.  Prisms  in  general.  Ex.15.    Truncated  prisms. 


BOOK  VIII 


CYLINDERS   AND   CONES 

CYLINDERS 

676.  A  cylindrical  surface  is 
a  curved  surface  generated  .by  a 
straight  line  which  moves  so  as 
constantly  to  touch  a  given  fixed 
curve  and  constantly  be  parallel 
to  a  given  fixed  straight  line. 

Thus,  every  shadow  cast  by  a  point 
of  light  at  a  great  distance,  as  by  a 
star  or  the  sun,  approximates  the 
cylindrical  form,  that  is,  is  bounded  Cylindrical  surface 

by  a  cylindrical  surface  of  light.  Hence,  in  all  radiations  (as  of  light, 
heat,  magnetism,  etc.)  from  a  point  at  a  great  distance,  we  are 
concerned  with  cylindrical  surfaces  and  solids. 

677.  The  generatrix  of  a  cylindrical  surface  is  the  mov- 
ing straight  line;  the  directrix  is  the  given  curve,  as  CDE\ 
an  element  of  the  cylindrical  surface  is  the  moving  straight 
line  in  any  one  of  its  positions,  as  DF. 

678.  A  cylinder  is  a  solid  bounded  by 
a  cylindrical  surface  and  by  two  parallel 
planes. 

The  bases  of  a  cylinder  are  its  parallel 
plane   faces;     the   lateral    surface    is    the 
cylindrical   surface   included    between  the 
parallel  planes  forming  its  bases;   the  alti-        cylinder 
tude  of  a  cylinder  is  the  distance  between  the  bases. 

The  elements  of  a  cylinder   are   the  elements   of   the 
cylindrical  surface  bounding  it. 

(403) 


404 


BOOK  VIII.     SOLID   GEOMETRY 


679.  Property  of  a  cylinder  inferred  immediately.    All 

the  elements  of  a  cylinder  are  equal,  for  they  are  parallel  lines 
included  between  parallel  planes  (Arts.  532,  676). 

The  cylinders  most  important  in  practical  life  are  those  determined 
by  their  stability,  the  ease  with  which  they  can  be  made  from  com- 
mon materials,  etc. 

680.  A  right  cylinder    is    a    cylinder 
whose   elements    are    perpendicular    to    the 


Oblique  circular 
cylinder 


681.  An  oblique  cylinder  is  one  whose 
elements  are  oblique  to  the  bases. 

682.  A  circular  cylinder  is  a   cylinder 
whose  bases  are  circles. 

683.  A  cylinder  of  revolution  is  a  cylin- 
der generated  by  the  revolution   of  a  rect- 
angle about  one  of  its  sides  as  an  axis. 

Hence,  a  cylinder  of  revolution  is  a  right 
circular  cylinder. 

Some  of  the  properties  of  this  solid  are  derived 
most  readily  by  considering  it  as  generated  by  a  re- 
volving rectangle  ;  and  others,  by  regarding  it  as  a 
particular  kind  of  cylinder  derived  from  the  general 
definition. 

684.  Similar  cylinders  of  revolution  are  cylinders  gen- 
erated   by    similar    rectangles    revolving    about    homologous 
sides. 

685.  A  tangent  plane   to   a  cylinder  is  a  plane  which 
contains  one  element  of  the  cylinder,  and  which  does  not  cut 
the  cylinder  on  being  produced. 


Ex.  1.  A  plane  passing  through  a  tangent  to  the  base  of  a  circu- 
lar cylinder  and  the  element  drawn  through  the  point  of  contact  is 
tangent  to  the  cylinder.  (For  if  it  is  not,  etc.) 

Ex.  2.  If  a  plane  is  tangent  to  a  circular  cylinder,  its  intersection 
the  itlane  of  the  base  is  tangent  to  the  base. 


CYLINDEKS  405 

686.  A  prism  inscribed  in  a  cylinder  is  a  prism  whose 
lateral  edges  are  elements  of  the  cylinder,  and  whose  bases 
are  polygons  inscribed  in  the  bases  of  the  cylinder. 


inscribed  prism  Circumscribed  prism 

687.  A  prism  circumscribed  about  a  cylinder  is  a  prism 
whose  lateral  faces  are  tangent  to  the  cylinder,  and  whose 
bases  are  polygons  circumscribed  about  the  bases   of  the 
cylinder. 

688.  A  section  of  a  cylinder  is  the  figure  formed  by  the 
intersection* of  the  cylinder  by  a  plane. 

A  right  section  of  a  cylinder  is  a  section  formed  by  a 
plane  perpendicular  to  the  elements  of  the  cylinder. 

689.  Properties  of  circular  cylinders.     By  Art.  441  the 
area  of  a  circle  is  the  limit  of  the  area  of  an  inscribed  or 
circumscribed  polygon,  and  the  circumference  is  the  limit 
of  the  perimeters  of  these  polygons;  hence 

1.  The  volume  of  a  circular  cylinder  is  the  limit  of  the 
volume  of  an  inscribed  or  circumscribed  prism. 

2.  The  lateral  area  of  a  circular  cylinder  is  the  limit  of 
the  lateral  area  of  an  inscribed  or  circumscribed  prism. 

Also,  3.  By  methods  loo  advanced  for  this  book,  it  may  be  proved  that 
the  perimeter  of  a  right  section  is  the  limit  of  the  perimeter  of  a  right  sec- 
tion of  an  inscribed  or  circumscribed  jpr  ism. 


406  BOOK  VIII.       SOLID     GEOMETKY 


PROPOSITION  I.     THEOREM 

690.    Every  section  of  a  cylinder  made  by  a  plane  pass- 
ing through  an  element  is  a  parallelogram. 


Given  the  cylinder  AQ  cut  by  a  plane  passing  through 
the  element  AB  and  forming  the  section  ABQP. 

To  prove  ABQP  a  ZZ7  . 

Proof.  AP\\BQ.  Art.  531. 

It  remains  to  prove  thatvP$  is  a  straight  line  ||  AB. 
Through  P  draw  a  line  in  the  cutting  plane  ||  AB. 
This  line  will  also  lie  in  the  cylindrical  surface.  Art.  676. 

.*.  this  line  must  coincide  with  PQ, 

(for  the  line  drawn  lies  in  both  the  cutting  plane  and  the  cylindrical 
surface,  hence,  it  must  be  their  intersection). 


:.  PQ  is  a --straight  line  1 1  AB. 

:.  ABQP  is  a  ZZ7  .  (Why  ?) 

Q.  E.  D. 

691.    COR.    Every  section  of  a  right  cylinder  made  by  a 
plane  passing  through  an  element  is  a  rectangle. 


Ex.  1.    A  door  swinging  on  its  hinges  generates  what  kind  of  a 
solid  f 

Ex.  2.    Every  section  of  a  parallelepiped  made  by  a  plane  inter- 
eectiug  all  its  lateral  edges  is  a  parallelogram, 


CYLINDEKS  407 


PROPOSITION  II.     THEOREM 
692.    The  bases  of  a  cylinder  are  equal. 


Given  the  cylinder  AQ  with  the  bases  APB  and  CQD. 

To  prove  base  APJ5  =  base  CQD. 

Proof.   Let  AC  and  BD  be  any  two  fixed  elements  in 
the  surface  of  the  cylinder  AQ. 

Take  P,  any  point  except  A  and  B  in  the  perimeter  of 
the  base,  and  through  it  draw  the  element  PQ. 

Draw  AB,  AP,  PB,  CD,  CQ,  QD. 

Then  AC  and  BD  are  =  and  ||.  (Why?) 

.-.  AD  is  a  ZZ7.  (Why?) 

Similarly  AQ  and  BQ  are  CU  . 

:.  AB=CD,  AP=CQ,  and  BP=DQ.         (Why?) 
/.  A  APB  =  A  CQD.  (Why  ?) 

Apply  the  base  APB  to  the  base  CQD  so  that  AB  coin- 
cides with  CD.     Then  P  will  coincide  with  Q, 
(for  A  APB  =  &  CQD). 

But  P  is  any  point  in  the  perimeter  of  the  base  APB. 

.*.  every  point  in  the  perimeter  of  the  lower  base  will 
coincide  with  a  corresponding  point  of  the  perimeter  of 
the  upper  base. 

.*.  the  bases  will  coincide  and  are  equal.  Art.  47. 

9.  £.  D. 


408  BOOK   VIII.       SOLID     GEOMETRY 

693.  COR.  1.   The  sections  of  a  cylinder  made  by  two 
parallel  planes  cutting  all  the  elements  arc  equal. 

For  the  sections  thus  formed  are  the  bases  of  the  cylinder 
included  between  the  cutting  planes. 

694.  COR.  2.  Any  section  of  a  cylinder  parallel  to  the 
base  is  equal  to  the  base. 


PROPOSITION  III.     THEOREM 

695.  The  lateral  area  of  a  circular  cylinder  is  equal  to 
the  product  of  the  perimeter  of  a  right  section  of  the  cylin- 
der by  an  element. 


Given  the  circular  cylinder  AJ,  having  its  lateral  area 
denoted  by  8,  an  element  by  E,  and  'the  perimeter  of  a 
right  section  by  P. 

To  prove  S=PXE. 

Proof.  Let  a  prism  with  a  regular  polygon  for  its  base 
be  inscribed  in  the  cylinder. 

Denote  the  lateral  area  of  the  inscribed  prism  by  $', 
and  the  perimeter  of  its  right  section  by  Pf. 

Then  the  lateral  edge  of  the  inscribed  prism  is  an  ele- 
ment of  the  cylinder.  Constr 


CYLINDERS  409 


Art.  608. 

If  the  number  of  lateral  faces  of  the  inscribed  prism  be 
indefinitely  increased, 

8'  will  approach  8  as  a  limit.  Art.  689,  2. 

P'  will  approach  P  as  a  limit.  Art.  689,  3. 

And  P'XE  will  approach  PXE  as  a  limit.  Art.  253,  2. 

But  S'  =  P'XE  always.  (Why?) 


/.   S=PXE.  (Why?) 

Q.  £.  D. 

696.  COR.  1.  The  lateral  area  of  a  cylinder  of  revolu- 
tion is  equal  to  the  product  of  the  circumference  of  its  base 
by  its  altitude. 

697.  Formulas  for  lateral  area  and  total  area  of  a  cylin- 
der of  revolution.    Denoting  the  lateral  area  of  a  cylinder  of 
revolution  by  8,  the  total  area  by  T,  the  radius  by  R,  and 
the  altitude  by  H. 

8=2  nRH. 

T=2  nRH  +  2  nR2  :.  T=2  nR  (H+  R). 


Ex.  1.  If,  in  a  cylinder  of  revolution,  H=W  in.  and  R=7  in., 
find  8  and  T. 

Ex.  2.  If  the  altitude  of  a  cylinder  of  revolution  equals  the  radius 
of  the  base  (H=R),  what  do  the  formulas  for  8  and  T  become  in 
terms  of  -K  ?  also,  in  terms  of  H  ? 

Ex.  3.  What  do  they  become,  if  the  altitude  equals  the  diameter 
of  the  base  ? 

Ex.  4.  In  a  cylinder  of  revolution,  what  is  the  ratio  of  the  lateral 
area  to  the  area  of  the  base  ?  to  the  total  area  ? 


410  BOOK   VIII.       SOLID     GEOMETRY 


PROPOSITION  IV.     THEOREM 

698.    The  volume  of  a  circular  cylinder  is  equal  to  the 
product  of  its  base  by  its  altitude. 


Given  the  circular  cylinder  AJ,  having  its  volume 
denoted  by  F,  its  base  by  B,  and  its  altitude  by  H. 

To  prove  V=BXH. 

Proof.  Let  a  prism  having  a  regular  polygon  for  its 
base  be  inscribed  in  the  cylinder,  and  denote  the  volume  of 
the  inscribed  prism  by  T7,  and  its  base  by  B' '. 

The   prism   will    have   the   same   altitude,    H,   as   the 

cylinder. 

.-.    V'  =  B'XH.  (Why?) 

If  the  number  of  lateral  faces  of  the  inscribed  prism 
be  indefinitely  increased, 

V  will  approach  V  as  a  limit.         Art.  689,  1. 

B'  will  approach  B  as  a  limit.  (Why?) 

And    B'  X  H  will  approach  B  X  H  as  a  limit       (Why  ?) 

But  V'  =  B'XH  always,  (Why  ?) 

/.    V=BXH.  (Why?) 

Q.  E.  D. 

699.  Formula  for  the  volume  of  a  circular  cylinder, 
By  use  of  Art.  450, 


CYLINDERS 

PROPOSITION  V.     THEOREM 


411 


700.  The  lateral  areas,  or  the  total  areas,  of  two  simi- 
lar cylinders  of  revolution  are  to  each  other  as  the  squares 
of  their  radii,  or  as  the  squares  of  their  altitudes;  and 
their  volumes  are  to  each  other  as  the  cubes  of  their  radii, 
or  as  the  cubes  of  their  altitudes. 


H' 


Given  two  similar  cylinders  of  revolution  having  their 
lateral  areas  denoted  by  S  and  $',  their  total  areas  by  T 
and  T',  their  volumes  by  Fand  V,  their  radii  by  R  and 
R',  and  their  altitudes  by  H  and  R' ,  respectively. 

To  prove  8  :  S'=T  :  T'  =  R2  :  R'2  =  R2  :  H'2-, 
and  V:  Vf  =  R*  :  Rf3  =  H3  :  H'3. 


g_^_  R+R 
R    R'    H'  +  R 


8'    2 


RXH     R  ^R 


(H+B)   =  B      H+B 
(R'  +  Rf)     R'    R' 


Also 


R' 


=R2     R  =R*  ^H3 
nR'2Rf  ~  R'2    H!    Rf3  ~  R'* 


Arts.  321,  3fflL 


R2 

W2 


R'2 

(Why?) 

(Why  ?) 
Q.  E.  D. 

Ex.  If  a  cylindrical  cistern  is  12  ft.  deep,  how  much  more  cement 
is  required  to  line  it  than  to  line  a  similar  cistern  6  ft.  deep  ?  How 
much  more  water  will  the  former  cistern  hold  ? 


412 


BOOK  VIII.      SOLID    GEOMETRY 


B 

Conical  surface 


CONES 

701.  A  conical  surface  is  a  sur- 
face   generated    by    a    straight    line 
which  moves  so  as  constantly  to  touch 
a  given  fixed  curve,  and  constantly 
pass  through  a  given  fixed  point. 

Thus  every  shadow  cast  by  a  near  point 
of  light  is  conical  in  form,  that  is,  is!  bounded 
by  a  conical  surface  of  light.  Hence,  the 
study  of  conical  surfaces  and  solids  is  im- 
portant from  the  fact  that  it  concerns  all 
cases  of  forces  radiating  from  a  near  point. 

702.  The  generatrix  of  a  conical   c\ 
surface  is  the  moving  straight  line,  as 
AAf;   the  directrix  is  the  given  fixed 

curve,  as  ABC-,    the  vertex  is  the  fixed  point,  as  0;    an 
element  is  the  generatrix  in  any  one  of  its  positions,  as  BB' '. 

703.  The  upper  and  lower  nappes  of  a  conical  surface 
are  the  portions  above    and    below    the  vertex,  respectively, 
as  0-ABC  and  0-A'B'C''. 

Usually  it  is  convenient  to  limit  a  conical  surface  to  a  single  nappe. 

704.  A  cone  is  a  solid  bounded  by  a 
conical  surface  and  a  plane  cutting  all  the 
elements. 

705.  The  base  of  a   cone  is  the  face 
formed   by  the  cutting  plane;   the  lateral 
surface  is  the  bounding  conical    surface; 
the  vertex  of  the  cone  is  the  vertex  of  the 
conical  surface;   the  elements  of  the  cone 
are  the  elements  of   the  conical  surface; 

the  altitude  of  a  cone  is  the  perpendicular  distance  from  the 
Vertex  to  the  plane  of  the  base. 


Oblique  circular 
cone 


CONES  413 

706.  A  circular  cone  is  a  cone  whose  base  is  a  circle. 
The  axis  of  &  circular  cone  is  the  line  drawn  from   the 
vertex  to  the  center  of  the  base. 

707.  A  right  circular  cone  is  a  circular  cone  whose  axis 
is  perpendicular  to  the  plane  of  the  base. 

An  oblique  circular  cone  is  a  circular  cone 
whose  axis  is  oblique  to  the  base. 


708.  A  cone  of  revolution  is  a  cone  gene- 
rated by  the  revolution  of  a  right  triangle 
about  one  of  its  legs  as  an  axis. 

Hence  a  cone  of  revolution  and  a  right 
circular  cone  are  the  same  solid.  Cone  of  revolution 

709.  Properties   of   a  cone   of   revolution   inferred  im- 
mediately. 

1.  The  altitude,  of  a  cone  of  revolution  is  the  axis  of  the 
cone. 

2.  All  the  elements  of  a  cone  of  revolution  are  equal. 

710.  The  slant  height  of  a  cone  of  revolution  is  any  one 

of  its  elements. 

711.  Similar  cones  of  revolution  are  cones  generated  by 
similar  right  triangles  revolving  about  homologous  sides. 

712.  A  plane  tangent  to  a  cone  is   a  plane  which  con- 
tains one  element  of  the  cone,  but  which  does  not  cut  the 
conical  surface  on  being  produced. 

Ex.  1.  A  plane  passing  through  a  tangent  to  the  base  of  a  circular     . 
cone  and  the  element  drawn  through  the  point  of  contact  is  tangent  to 
the  cone. 

Ex.  2.  If  a  plane  is  tangent  to  a  circular  cone,  its  intersection  with 
the  plane  of  the  base  is  tangent  to  the  cone. 


414 


BOOK  VTTT.      SOLID    GEOMETRY 


713.  A  pyramid  inscribed  in 
a  cone  is  a  pyramid  whose  lateral 
edges  are  elements  of  the  cone 
and  whose  base  is  a  polygon  in- 
scribed in  the  base  of  the  cone. 

714.  A   pyramid    circum- 
scribed about  a  cone  is  a  pyra- 
mid whose  lateral  faces  are  tan 

gent  to  the  cone  and  whose  base  is  a  polygon  circumscribed 
about  the  base  of  the  cone. 

715.  Properties  of  circular  cones.    By  Art.  441  the  area 
of  a  circle  is  the  limit  of  the  area  of  an  inscribed,  or  of  a 
circumscribed  polygon,  and  the  circumference  is  the  limit 
of  the  perimeters  of  these  polygons;  hence 

1.  The  volume  of  a  circular  cone  is  the  limit  of  the  vol- 
ume of  an  inscribed  or  circumscribed  pyramid. 

2.  The  lateral  area  of  a  circular  cone  is  the  limit  of  the 
lateral  area  of  an  inscribed  or  circumscribed  pyramid. 


716.  A  frustum  of  a  cone  is  the  por- 
tion of  the  cone  included  between  the  base 
of  the  cone  and  a  plane  parallel  to  the 
base. 

The  lower  base  of  the  frustum  is  the 
base  of  the  cone,  and  the  upper  base  of 
the  frustum  is  the  section  made  by  the 
plane  parallel  to  the  base  of  the  cone. 


Frustum  of  a  cone. 


What  must  be  the  altitude  and  the  lateral  surface  of  a 
frustum  of  a  cone;  also  the  slant  height  of  the  frustum  of 
a  cone  of  revolution  ? 


CONES 


415 


X        PROPOSITION  VI.    THEOREM 

717.    Every  section  of  a  cone  made  by  a  plane  passing 
through  its  vertex  is  a  triangle  ^^  &£  Y^vv-vr^          *»*•*•  <<J2 


B 


Given  the  cone  S-APBQ  with  a  plane  passing  through 
the  vertex  S9  and  making  the  section  SPQ. 

To  prove  SPQ  a  triangle. 

Proof.   PQ,  the  intersection  of  the  base  and  the  cutting 

plane,  is  a  straight  line.  , 

r/°v<i  ^*t*^~ffrf"  &**- *-  6-~— r»-~ 

Draw  the  straight  lines  SP  and  SQ. 

Then  SP  and  /SQ  must  be  in  the  cutting  plane ;  Art.  498. 

And  be  elements  of  the  conical  surface.  Art.  701. 

.'.  the  straight  lines  8P  and  SQ  are  the  intersections  of 
the  conical  surface  and  the  cutting  plane. 


/.  the  section  SPQ  is  a  triangle, 

(for  it  is  bounded  by  three  straight  lines). 


Art.  81. 


Q.  £.  D. 


Ex.    What  kind  of  triangle  is  a  section  of  a  right  circular  cone 
made  by  a  plane  through  the  vertex  ? 


416 


BOOK   VIII.       SOLID     GEOMETKY 


PROPOSITION  VII.    THEOREM 

718.    Every  section  of  a  circular  cone  made  by  a  plane 
parallel  to  the  base  is  a  circle. 


Given  the  circular  cone  SAB  with  apb  a  section  made 
by  a  plane  parallel  to  the  base. 

To  prove  apb  a  circle. 

Proof.  Denote  the  center  of  the  base  by  0,  and  draw 
the  axis,  SO,  piercing  the  plane  of  the  section  in  o. 

Through  SO  and  any  element,  SP,  of  the  conical  sur- 
face, pass  a  plane  cutting  the  plane  of  the  base  in  the 
radius  OP,  and  the  plane  of  the  section  in  op. 

In  like  manner,  pass  a  plane  through  SO  and  SB  form- 
ing the  intersections  OB  and  ob. 

:.  OP\\op,  and  OB  \\  ob.  (Why?) 

.'.  A  SPO  and  SBO  are  similar  to  A  Spo  and  Sbo, 
respectively. 


But 


op  _     So_     _  °b 

OP~  \soj  ~  OB' 
OP=  OB. 

.*.  op  =  ob. 
:.  opb  is  a  circle. 


Art.  328. 
(Why?) 


(Why?) 

(Why?) 

(Why?) 
Q.  E.  D. 

719.    COR.    The  axis  of  a  circular  cone  passes  through 
the  center  of  every  section  parallel  to  the  base. 


CONES  417 

PROPOSITION  VIII.    THEOREM 

720.  The  lateral  area  of  a  cone  of  revolution  is  equal  to 
half  the  product  of  the  slant  height  by  the  circumference  of 
the  base. 


Given  a  cone  of  revolution  having  its  lateral  area  de- 
noted by  8,  its  slant  height  by  L,  and  the  circumference 
of  its  base  by  (7. 

To  prove  #=J  G  X  L. 

Proof.  Let  a  regular  pyramid  be  circumscribed  about 
the  cone. 

Denote  the  lateral  area  of  the  pyramid  by  $',  and  the 
perimeter  of  its  base  by  P. 

Then  8'  =  J  P  X  L.  Art.  642. 

If  the  number  of  lateral  faces  of  the  circumscribed 
pyramid  be  indefinitely  increased,  . 

8'  will  approach  8  as  a  limit.          Art.  715,  2. 
P  will  approach  C  as  a  limit.  Art.  441. 

And  J  P  XL  will  approach  J  CXL  as  a  limit.   Art.  253,  2. 

But  8'  =  %  P  X  L  always.  (Why  ?) 

/.   $=J  CXL.  (Why?) 

Q.  E.  D. 

721.  Formulas  for  lateral  area  and  total  area  of  a  cone 
of  revolution.  Denoting  the  radius  of  the  base  by  R, 

8= i  (2  TiR  X  L)  :.  S=nRL. 
Also       T=  nRL  +  nR2  .'.  T=nR(L  +  R). 

AA 


418 


BOOK   VIII.       SOLID     GEOMETRY 


PROPOSITION  IX.     THEOREM 


722.    The  volume  of  a  circular  cone  is  equal  to  one-third 
of  the  product  of  its  base  by  its  altitude. 


Given  a  circular  cone  having  its  volume  denoted  by  V, 
its  base  by  B,  and  its  altitude  by  H. 

To  prove  V=  J  B  X  H. 

Proof.    Let  a  pyramid  with  a  regular  polygon  for  its 
base  be  inscribed  in  the  given  cone. 

Denote  the  volume  of  the  inscribed  pyramid  by  V ',  and 
its  base  by  B' . 

Hence  V  =  J  B'  X  H.  Art.  esi. 

If  the  number  of  lateral  faces  of  the  inscribed  pyramid 
be  indefinitely  increased, 

V  will  approach  V  as  a  limit.  (Why?) 

B'  will  approach  B  as  a  limit.  (Why?) 

And  i  Bf  X  H  will  approach  J  B  X  H  as  a  limit.  (Why  ?) 

But  V  =  $  B'X  H  always.  (Why  ?) 

/.  F=i£  X  If.  (Why?) 

Q.  E.  D. 

723.   Formula  for  the  volume  of  a  circular  cone. 


Ex.  1.    If,  in  a  cone  of  revolution,  H=3  and  .R=4,  find  £,  T  and  F. 

Ex.  2.    If  the  altitude  of  a  cone  of  revolution  equals  the  radius  of 
the  base,  what  do  the  formulas  for  S,  T  and  V  become  ? 


CONES  419 

PROPOSITION  X.     THEOREM 

724.  The  lateral  areas,  or  the  total  areas,  of  two  simi- 
lar cones  of  revolution  are  to  each  other  as  the  squares  of 
their  radii,  or  as  the  squares  of  their  altitudes,  or  as  the 
squares  of  their  slant  heights;  and  their  volumes  are  to  each 
other  as  the  cubes  of  these  lines. 


Given  two  similar  cones  of  revolution  having  their 
lateral  areas  denoted  by  8  and  $',  their  total  areas  by  T 
and  T' ,  their  volumes  by  V  and  V,  their  radii  by  R  and 
Rf,  their  altitudes  by  R  and  H',  and  their  slant  heights  by 
L  and  U ',  respectively. 

To  prove  8:8'=T:  T'  =  ft2:  R'2  =  H2  :  H'2  =  L2  :  £'2; 
and  V:  V'  = 

T?        T.         T.  4-  7? 

(Why?) 


=  ^       =       = 

~  ~       ~ 


#'    TiE'l/    J?    J7    £/2    L12    H'2 

==TT,X' 


T'     TCE'dX  +  .R')     R1    L'  +  Rf    R'2    L'2    H'2 

(Why?) 

^  F=4  n&H!  =  &**H'  =  W  =  H*  =  lfi'     (Why?) 

Q.  E.  D. 

725.  DEP.  An  equilateral  cone  is  a  cone  of  revolution 
such  that  a  section  through  the  axis  is  an  equilateral 
triangle. 


420  BOOK  VIII.       SOLID     GEOMETRY 

y    PROPOSITION  XL     THEOREM 

726.  The  lateral  area  of  a  frustum  of  a  cone  of  revolu- 
tion is  equal  to  one-half  the  sum  of  the  circumferences  of  its 
bases  multiplied  by  its  slant  height. 


Given  a  frustum  of  a  cone  of  revolution  having  its 
lateral  area  denoted  by  8,  its  slant  height  by  L,  the  radii 
of  its  bases  by  R  and  r,  and  the  circumferences  of  its  bases 
by  C  and  c. 

To  prove  fl=J  (C+c)XL. 

Proof.  Let  the  frustum  of  a  regular  pyramid  be  circum- 
scribed about  the  given  frustum.  Denote  the  lateral  area 
of  the  circumscribed  frustum  by  S',  the  perimeter  of  the 
lower  base  by  P,  and  the  perimeter  of  the  upper  base  by  p. 

The  slant  height  of  the  circumscribed  frustum  is  L. 

Hence  8'  =  %  (P  +  p)  X  L.  Art.  643. 

Let  the  pupil  complete  the  proof. 

Q.  E.  D. 

727.  Formula  for  the  lateral  area  of  a  frustum  of  a  cone 
of  revolution.        £=J  (2  nR-\-  2  nr)  L. 

:.  S=n  (B  +  r)L. 

728.  COR.   The  lateral  area  of  a  frustum  of  a  cone  of 
revolution  is  equal  to  the  product  of  the  circumference  of  its 
rtiidsection  by  its  slant  heiyht. 


CONES  421 

>^  PROPOSITION  XII.     THEOREM 

729.  The  volume  of  Hie  frustum  of  a  circular  cone  is 
equivalent  to  the  volume  of  three  cones,  whose  common  alti- 
tude is  the  altitude  of  the  frustum,  and  whose  bases  are  the 
lower  base,  the  upper  base,  and  a  mean  proportional  between 
the  two  bases. 


Given  a  frustum  of  a  circular  cone  having  its  volume 
denoted  by  F,  its  altitude  by  H,  the  area  of  its  lower  base 
by  B,  and  that  of  its  upper  base  by  b. 

To  prove          F  =  4  H  (B  +  b  +  l/BXb). 

Proof.  Let  the  frustum  of  a  pyramid  with  regular  poly- 
gons for  its  bases  be  inscribed  in  the  given  frustum. 
Denote  the  volume  of  the  inscribed  frustum  by  F',  and  the 
areas  of  its  bases  by  B'  and  bf. 

':.  F  =  J  H  (Bf  +  b'  +  VB*  X  b') .  (Why?) 

If  the  number  of  lateral  faces  of  the  inscribed  frustum 
be  indefinitely  increased,  V  will  approach  F,  Br  and  bf 
approach  B  and  b  respectively,  and  Bf  X  b'  approach  B  X 
5,  as  limits.  Art.  715. 

Hence,  also,  B'  +  b'+T/B'  X  b'  will  approach  B  +  b  + 

V B  X  6  as  a  limit.  Art.  253. 

But  7/=  J  H  (Br  +  V  +  *JB'  X  V)  always.  (Why  t) 

/.  V=  J  H  (B  +  b  +  l/BXb).  (Why  f ) 

Q.  E.  J>. 


422  BOOK  VIII.       SOLID     GEOMETKY 

730.    Formula  for  the  volume  of  the  frustum  of  a  circu- 
lar cone. 


F=4  H  (nR2  +  nr2  +     n&  X  nr2)  . 


Ex.  1.  The  measurement  of  the  volume  of  a  frustum  of  a  cone  of 
revolution  reduces  to  the  measurement  of  the  lengths  of  what  straight 
lines? 

Ex.  2.  If  a  conical  oil -can  is  12  in.  high,  how  much  more  tin  is 
required  to  make  it  than  to  make  a  similar  oil -can  6  in.  high  ?  How 
much  more  oil  will  it  hold  ? 

Ex.  3.  The  linear  dimensions  of  a  conical  funnel  are  three  times 
those  of  a  similar  funnel.  How  much  more  tin  is  required  to  make 
the  first  ?  How  much  more  liquid  will  it  hold  ? 

Ex.  4.  Make  a  similar  comparison  of  cylindrical  oil -tanks.  Of 
conical  canvas  tents. 

EXERCISES.     CROUP  7O 

THEOREMS    CONCERNING    CYLINDERS    AND    CONES 

Ex.  1.  Any  section  of  a  cylinder  of  revolution  through  its  axis  is 
a  rectangle. 

Ex.  2.  On  a  cylindrical  surface  only  one  straight  line  can  be 
drawn  through  a  given  point. 

[SuG.    For  if  two  straight  lines  could  be  drawn,  etc.] 

Ex.  3.  The  intersection  of  two  planes  tangent  to  a  cone  is  a 
straight  line  through  the  vertex. 

Ex.  4.  If  two  planes  are  tangent  to  a  cylinder,  their  line  of  inter- 
section is  parallel  to  an  element  of  the  cylinder. 

[Sue.   Pass  a  plane  J_  to  the  elements  of  the  cylinder.] 

Ex.  5.  If  tangent  planes  be  passed  through  two  diametrically 
opposite  elements  of  a  circular  cone,  these  planes  intersect  in  a 
straight  line  through  the  vertex  and  parallel  to  the  plane  of  the  base. 

Ex.  6.  In  a  cylinder  of  revolution  the  diameter  of  whose  base 
equals  the  altitude,  the  volume  equals  one-third  the  product  of  the 
total  surface  by  the  radius  of  the  base. 


EXEECISES  ON  THE  CYLINDER  AND  CONE   423 


Ex.  7.  A  cylinder  and  a  cone  of  revolution  have  the  same  base 
and  the  same  altitude.  Find  the  ratio  of  their  lateral  surfaces,  and  also 
of  their  volumes. 

Ex.  8.  If  an  equilateral  triangle  whose  side  is  a  be  revolved  about 
one  of  its  sides  as  an  axis,  find  the  area  generated  in  terms  of  a. 

Ex.  9.  If  a  rectangle  whose  sides  are  a  and  &  be  revolved  first 
about  the  side  a  as  an  axis,  and  then  about  the  side  &,  find  the  ratio  of 
the  lateral  areas  generated,  and  also  of  the  volumes. 

Ex.  10.  The  bases  of  a  cylinder  and  of  a  cone  of  revolution  are 
concentric.  The  two  solids  have  the  same  altitude,  and  the  diameter 
of  the  base  of  the  cone  is  twice  the  diameter  of  the  base  of  the  cylin- 
der. What  kind  of  line  is  the  intersection  of  their  lateral  surfaces, 
and  how  far  is  it  from  the  base  ? 

Ex.  11.  Determine  the  same  when  the  radius  of  the  cone  is  three 
times  the  radius  of  the  cylinder.  Also  when  r  times. 

Ex.  12.  Obtain  a  formula  in  terms  of  r  for  the  volume  of  the 
frustum  of  an  equilateral  cone,  in  which  the  radius  of  the  upper  base 
is  r  and  that  of  the  lower  base  is  3r. 

Ex.  13.  A  regular  hexagon  whose  side  is  a  revolves  about  a  diag- 
onal through  the  center  as  axis.  Find,  in  terms  of  a,  the  surface 
and  volume  generated. 

Ex.  14.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given 
straight  line. 

Ex.  15.  Find  the  locus  of  a  point  whose  distance  from  a  given 
line  is  in  a  given  ratio  to  its  distance  from  a  fixed  plane  perpen- 
dicular to  the  line. 

Ex.  16.  Find  the  locus  of  all  straight  lines  which  make  a  given 
angle  with  a  given  line  at  a  given  point. 

Ex.  17.  Find  the  locus  of  all  straight  lines  which  make  a  given 
angle  with  a  given  plane  at  a  given  point. 

Ex.  18.    Find  the  locus  of  all  points  at  a  given  dis- 
tance from  the  surface  of  a  given  cylinder  of  revolution. 

Ex.  19.    Find  the  locus  of  all  points  at  a  given  dis- 
tance from  the  surface  of  a  given  cone  of  revolution. 


424  BOOK   VIII.       SOLID     GEOMETRY 

EXERCISES.     CROUP  71 

PROBLEMS    CONCERNING    THE    CYLINDER    AND    CONE 

Ex.  1.  Through  a  given  element  of  a  circular  cylinder,  pass  a 
plane  tangent  to  the  cylinder. 

Ex.  2.  Through  a  given  element  of  a  circular  cone,  pass  a  plane  tangc  pt 
to  the  cone. 

Ex.  3.  About  a  given  circular  cylinder  circumscribe  a  prism,  with 
a  regular  polygon  for  its  base. 

Ex.  4.  Through  a  given  point  outside  a  circular  cylinder,  pass  a 
plane  tangent  to  the  cylinder. 

Ex.  5.  Through  a  given  point  outside  a  given  circular  cone,  pass 
a  plane  tangent  to  the  cone. 

[Suo.  Through  the  vertex  of  the  cone  and  the  given  point  pass  a 
line,  and  produce  it  to  meet  the  plane  of  the  base.] 

Ex.  6.  Into  what  segments  must  the  altitude  of  a  cone  of  revolution 
be  divided  by  a  plane  parallel  to  the  base,  in  order  that  the  volume  of 
the  cone  be  bisected  ? 

Ex.  7.  Divide  the  lateral  surface  of  a  given  cone  of  revolution  into 
two  equivalent  parts  by  a  plane  parallel  to  the  base. 

Ex.  8.  If  the  lateral  surface  of  a  cylinder  of  revolution  be  cut 
along  one  element  and  unrolled,  what  sort  of  a  plane  figure  is  formed  ? 

Hence,  out  of  cardboard  construct  a  cylinder  of  revolution  with 
given  altitude  and  given  circumference. 

Ex.  9.  If  the  lateral  surface  of  a  cone  of .  revolution  be  cut  along 
one  element  and  unrolled,  what  sort  of  a  plane  figure  is  formed  ? 

Hence,  out  of  cardboard  construct  a  cone  of  revolution  of  given 
slant  height. 

Ex.  10.    Construct  an  equilateral  cone  out  of  pasteboard. 

Ex.  11.  Construct  a  frustum  of  a  cone  of  revolution  out  of  paste- 
board. 


BOOK    IX 

THE    SPHERE 

731.  A  sphere  is  a  solid  bounded  by  a  surface  all 
points  of  which  are  equally  distant  from  a  point  within 
called  the  center. 


732.  A  sphere  may  also  be  defined  as  a  solid  generated 
by  the  revolution  of  a  semicircle  about  its  diameter  as  an 
axis. 

.   Some  of  the  properties  of  a  sphere  may  be  obtained  more  readily 
from  one  of  the  two  definitions  given,  and  some  from  the  other. 

A  sphere  is  named  by  naming  the  point  at  its  center,  or  by  naming 
three  or  more  points  on  its  surface. 

733.  A  radius  of  a  sphere  is  a  line  drawn  from  the 
center  to  any  point  on  the  surface. 

A  diameter  of  a  sphere  is  a  line  drawn  through  the 
center  and  terminated  at  each  end  by  the  surface  of  the 
sphere. 

734.  A  line  tangent  to  a  sphere  is  a  line  having  but  one 
point  in  common  with  the  surface  of  the  sphere,  however 
far  the  line  be  produced. 

(425) 


426 


BObK  IX.      SOLID     GEOMETKY 


735.  A  plane  tangent  to  a  sphere  is  a  plane  having  but 
one  point  in  common  with  the  surface  of  the  sphere,  how- 
ever far  the  plane  be  produced. 

736.  Two  spheres  tangent  to  each  other  are  spheres 
whose  surfaces  have  one  point,  and  only  one,  in  common. 

737.  Properties  of  a  sphere  inferred  immediately. 

1.  All  radii  of  a  sphere,  or  of  equal  spheres,  are  equal. 

2.  All  diameters  of  a  sphere,   or  of  equal  spheres,  are 
equal. 

3.  Two  spheres  are  equal  if  their  radii  or  their  diame- 
ters are  equal. 


PROPOSITION  I.     THEOREM 
738.    A  section  of  a  sphere  made  by  a  plane  is  a  .circle. 


Given  the  sphere  0,  and  PCD  a  section  made  by  a  plane 
cutting  the  sphere. 

To  prove  that  PCD  is  a  circle. 

Proof.  From  the  center  0,  draw  OA  J_  the  plane  of 
the  section. 


THE     SPHERE  427 

Let  0  be  a  fixed  point  on  the  perimeter  of  the  section, 
and  P  any  other  point  on  this  perimeter. 
Draw  A  C,  AP,  OC,  OP. 
Then  the  A  GAP  and  OAC  are  rt.  A.  Art.  505. 

OP=  OC.  (Why  ?) 

OA=OA.  (Why?) 

A0AP=AOJLO.  (Why?) 

.'.  AP=AC.  (Why?) 

But  P  is  any  point  on  the  perimeter  of  the  section  PCD. 
.'.  every  point  on  this  perimeter  is  at  the  distance  AC 
from  A. 

:.  PCD  is  a  circle  with  center  A.  Art.  197. 

Q.  E.  D. 

739.  COR.  1.    Circles  which   are  sections  of  a   sphere 
made  by  planes  equidistant  from  the  center  are  equal;    and 
conversely. 

740.  COR.  2.    Of  two  circles  on  a  sphere,  the  one  made 
by  a  plane  more  remote  from  the  center  is  smaller;    and 
conversely. 

741.  DEF.    A  great  circle  of  a  sphere  is  a  circle  whose 
plane  passes  through  the  center  of  the  sphere. 

742.  DEF.    A  small  circle  of  a  sphere  is  a  circle  whose 
plane  does  not  pass  through  the  center  of  the  sphere. 

743.  DEF.    The  axis  of  a  circle   of  a  sphere   is  the 

diameter  of  a  sphere  which  is  perpendicular  to  the  plane  of 
the  circle.     Thus,  on  figure  p.  426,  BB'  is  the  axis  of  PCD. 

744.  DEF.    The  poles  of  a  circle  of  a  sphere  are  the 
extremities  of  the  axis  of  the  circle.     Thus,  B  and  B',  of 
figure  p.  426,  are  poles  of  the  circle  PCD. 


428  BOOK  IX.      SOLID    GEOMETRY 

745.     Properties  of  circles  of  a  sphere  inferred  imme- 
diately. 

1.  The  axis  of  a  circle  of  a  sphere  passes  through  the 
center  of  the  circle;  and  conversely. 

2.  Parallel  circles  have  the  same  axis  and  the  same  poles. 

3.  All  great  circles  of  a  sphere  are  equal. 

1  great  circle  on  a  sphere  bisects  the  sphere  and 


its  surface. 

5.  Two  great  circles  on  a  sphere  bisect  each  other. 

For  the  line  of  intersection  of  the  two  planes  of  the 
circles  passes  through  the  center,  and  hence  is  a  diameter 
of  each  circle. 

6.  Through  two  points  (not  the  extremities  of  a  diameter) 
on  the  surface  of  a  sphere,  one,  and  only  one,  great  circle  can 
be  passed. 

For  the  plane  of  the  great  circle  must  also  pass  through 
the  center  of  the  sphere  (Art.  741),  and  through  three 
points  not  in  a  straight  line  only  one  plane  can  be  passed 
(Art.  500). 

7.  Through  any  three  points  on  the  surface  of  a  sphere, 
not  in  the  same  plane  with  the  center,  one  small  circle,  and 
only  one,  can  be  passed. 

746.  DEF.  The  distance  between  two  points  on  the  sur- 
face of  a  sphere  is  the  length  of  the  minor  arc  of  a  great 
circle  joining  the  points. 


Ex.  I.  If  the  radius  of  a  sphere  is  13  in.,  find  the  radius  of  a 
circle  on  the  sphere  made  by  a  plane  at  a  distance  of  1  ft.  from  the 
center. 

Ex.  2.  What  geographical  circles  on  the  earth's  surface  are  great, 
and  what  small  circles  ? 

Ex.  3.  What  is  the  largest  number  of  points  in  which  two  circles 
on  the  surface  of  a  sphere  can  intersect  ?  Why  ? 


THE     SPHERE 


429 


PROPOSITION  II.     THEOREM 

747.    All  points  in  the  circumference  of  a  circle  of  a 
sphere  are  equally  distant  from  each  pole  of  the  circle. 


Given  ABC  a  circle  of  a  sphere,  and  P  and  Pf  its  poles. 

To  prove  the  arcs  PA,  PB,  PC  equal,  and  arcs  P'A, 
P'B,  P'G  equal. 

Proof.   Draw  the  chords  PA,  PB,  PC. 

The  chords  PA,  PB  and  PC  are  equal.  Art.  518. 

/.  arcs  PA,  PB  and  PC  are  equal.  Art.  218. 

In  like  manner,  the  arcs  P'A,  P'B  and  P'C  may  be 
proved  equal. 

Q.  £.  D. 

748.  DEF.    The  polar  distance  of  a  small  circle  on  a 
sphere  is  the  distance  of  any  point  on  the  circumference  of 
the  circle  from  the  nearer  pole. 

The  polar  distance  of  a  great  circle  on  a  sphere  is  the  dis- 
tance of  any  point  on  the  circumference  of  the  great  circle 
from  either  pole. 

749.  COR.    The  polar  distance  of  a  great  circle  is  the 
quadrant  of  a  great  circle. 


430  BOOK  IX.      SOLID    GEOMETRY 

PROPOSITION  III.    THEOREM 

750.  If  a  point  on  the  surface  of  a  sphere  is  at  a  quad- 
rant's distance  from  two  other  points  on  the  surface,  it  is 
the  pole  of  the  great  circle  through  those  points. 


Given  PB  and  PC  quadrants   on   the  surface   of  the 
sphere  0,  and  ABC  a  great  circle  through  B  and  C. 

To  prove  that  P  is  the  pole  of  ABC. 
Proof.    From  the  center  0  draw  the  radii  OB,  OC,  OP. 
The  arcs  PB  and  PC  are  quadrants.  (Why  ?) 

.'.  A  POB  and  POC  are  rt.  A  .  (Why  ?) 

/.  PO  I.  plane  ABC.  (Why?) 

.'.  P  is  the  pole  of  the  great  circle  ABC.       (Why  ?) 

Q.  £.  D. 

751.    COR.    Through  two  given  points  on  the  surface  of 
a  sphere  to  describe  a  great  circle. 

Let  A  and  B  be  the  given  points. 
From  A  and  B  as  centers,  with  a 
quadrant  as  radius,  describe  arcs 
on  the  surface  of  the  sphere  inter- 
secting at  P.  With  P  as  a  center 
and  a  quadrant  as  a  radius,  describe  a  great  circle. 


THE     SPHERE  431 

• 

PROPOSITION  IV.     THEOREM 

752.    A  plane  perpendicular  to  a  radius  at  its  extremity 
is  tangent  to  the  sphere. 


Given  the  sphere  0,  and  the  plane  MN  J.  the  radius  OA 
of  the  sphere  at  its  extremity  A. 

To  prove  MN  tangent  to  the  sphere. 
Proof.    Take  P  any  point  in  plane  MN  except  A.    Draw 
OP.     Then  OP  >  OA.  (Why?) 

.*.  the  point  P  is  outside  the  surface  of  the  sphere. 
But  P  is  any  point  in  the  plane  MN  except  A. 
.'.plane  MN  is  tangent  to  the  sphere  at  the  point  A, 
(for  every  paint  in  the  plane,  except  A,  is  outside  the  surface  of  the  sphere). 

Art.  735.        Q.  E.  D. 

753.  COR.  1.  A  plane,  or  a  line,  which  is  tangent  to  a 
sphere,  is  perpendicular  to  the  radius  drawn  to  the  point  of 
contact.     Also,  if  a  plane  is  tangent  to  a  sphere,  a  perpendicu- 
lar to  the  plane  at  its  point  of  contact  passes  through  the  center 
of  the  sphere. 

754.  COR.  2.    A  straight  line  perpendicular  to  a  radius 
of  a  sphere  at  its  extremity  is  tangent  to  the  sphere. 

755.  COR.  3.    A  straight  line  tangent  to  a  circle  of  a 
sphere  lies  in  the  plane  tangent  to  the  sphere  at  the  point  of 
contact. 


432  BOOK  IX.       SOLID     GEOMETKY 

756.  COR.  4.   A  straight  line  drawn  in  a  tangent  plane, 
and  through  the  point  of  contact  is  tangent  to  the  sphere  at 
that  point. 

757.  COR.  5.    Two  straight  lines  tangent  to  a  sphere  at 
a  given  point  determine  the  tangent  plane  at  that  point. 

758.  DEP.    A  sphere  circumscribed  about  a  polyhedron 

is  a  sphere  in  whose  surface  lie  all  the  vertices  of  the 
polyhedron. 

759.  DEF.    A  sphere  inscribed   in  a  polyhedron   is   a 
sphere  to  which  all  the  faces  of  the  polyhedron  are  tangent. 

PROPOSITION  V.     PROBLEM 

760.  To   circumscribe   a   sphere   about   a  given    tetra- 
hedron. 


Given  the  tetrahedron  ABCD. 

To  circumscribe  a  sphere  about  ABCD. 

Construction  and  Proof.  Construct  E  and  F  the  centers 
of  circles  circumscribed  about  the  A  ABC  and  BCD,  re- 
spectively. Art.  286. 

Draw  EH  _L  plane  ABC  and  FK  _L  plane  BCD.  Art,  514. 

Draw  EQ  and  FG  to  <?  the  midpoint  of 


THE     SPHERE  433' 

Then  EG  and  FG  are  JL  EG.  Art.  113. 

/.  plane  EGF  _L  BC.  Art.  509. 

/.  plane  EGF  JL  plane  JU30.  Art.  555. 

/.  EH  lies  in  the  plane  FGE.  Art.  558. 
In  like  manner  FK  lies  in  the  plane  FGE. 

The  lines  EG  and  .F&  are  not  ||, 

(/or  Mey  TweeZ  in  the  point  G). 
:.  the  lines  EH  and  JPfiT  are  not  j|.  Art.  122. 

Hence  EH  must  meet  FK  in  some  point  0. 

But  EH  is  the  locus  of  all  points  equidistant  from  A, 
B  and  (7;  and  -FZT  is  the  locus  of  all  points  equidistant 
from  B,  G  and  D.  Art.  520. 

/.  0,  which  is  in  both  EH  and  FK,  is  equidistant  from 
A,  B,  CandD.  (Why?) 

Hence  a  spherical  surface  constructed  with  0  as  a  center 
and  OA  as  a  radius  will  pass  through  A,  B,  C  and  D,  and 
form  the  sphere  required.  Q.  E.  r. 

761.  COB.  1.  Four  points  not  in  the  same  plane  deter- 
mine a  sphere. 

762.  COR.  2.  The  four  perpendiculars   erected  at   the 
centers  of  the  faces  of  a  tetrahedron  meet  in  a  point. 

763.  COR.  3.   The  six  planes  perpendicular  to  the  edges 
of  a  tetrahedron  at  their  midpoints  intersect  in  a  point. 

764.  DEF.    An  angle  formed  by  two  curves  is  the  angle 
formed  by  a  tangent  to  each  curve  at  the  point  of  inter- 
section. 

765.  DEF.    A  spherical  angle  is  an  angle  formed   by 
two  intersecting  arcs  of  great  circles  on  a  sphere,   and 
hence  by  tangents  to  these  arcs  at  the  point  of  intersection. 

BB 


434  BOOK   IX.      SOLID     GEOMETRY 

PROPOSITION  VI.     PROBLEM 
766.    To  inscribe  a  sphere  in  a  given  tetrahedron. 


Given  the  tetrahedron  ABGD. 

To  inscribe  a  sphere  in  ABCD. 

Construction  and  Proof.  Bisect  the  dihedral  angle  D- 
A B- C  by  the  plane  GAB;  similarly  bisect  the  dihedral  A 
whose  edges  are  BG  and  AG  by  the  planes  GBG  and  GAG, 
respectively. 

Denote  the  point  in  which  the  three  bisecting  planes 
intersect  by  0. 

Every  point  in  the  plane  GAB  is  equidistant  from  the 
faces  DAB  and  GAB.  .  Art.  562. 

Similarly,  every  point  in  GBG  is  equidistant  from  the 
two  faces  intersecting  in  BG,  and  every  point  in  GAG  is 
equidistant  from  the  two  faces  intersecting  in  AG. 

:.  0  is  equidistant  from  all  four  faces  of  the  tetra- 
hedron. Ax.  1. 

Hence,  from  0  as  a  center,  with  the  JL  from  0  to  any 
one  face  as  a  radius,  describe  a  sphere. 

This  sphere  will  be  tangent  to  the  four  faces  of  the 
tetrahedron  and  .'.  inscribed  in  the  tetrahedron.  Art.  759. 

Q.  E.  F. 

767.  COR.  The  planes  bisecting  the  six  dihedral  angles 
of  a  tetrahedron  meet  in  one  point. 


THE     SPHEKE  435 

PROPOSITION  VII.    PROBLEM 
768.    To  find  the  radius  of  a  given  material  sphere. 

p 
{\ 

:^B  I    \  \  *j ~V 

'         *--:>-  jjs 

\      s^\  |   / 

2*~ '''  .  P' 

Fig-.  2  ^8'  3 

Given  the  material  sphere  0. 

To  construct  the  radius  of  the  sphere. 

Construction.  With  any  point  P  (Fig.  1)  of  the  surface 
of  the  sphere  as  a  pole,  describe  any  convenient  circum- 
ference on  the  surface. 

On  this  circumference  take  any  three  points  A,  B  and  C. 

Construct  the  A  ABC  (Fig.  2)  having  as  sides  the  three 
chords  AB,  BC,  AC,  obtained  from  Fig.  1,  by  use  of  the 
compasses.  Art.  283. 

Circumscribe  a  circle  about  the  A  ABC.  Art.  286. 

Let  KB  be  the  radius  of  this  circle. 

Construct  (Fig.  3)  the  right  A  Jcpb,  having  for  hypot- 
enuse the  chord  pb  (Fig.  1)  and  the  base  kb.  Art.  284. 

Draw  bp'  J_  bp  and  meeting  pk  produced  at  p' . 
Bisect  ppf  at  0. 

Then  op  is  the  radius  of  the  given  sphere. 
Proof.   Let  the  pupil  supply  the  proof, 

Q.  E.  F. 


436  BOOK   IX.      SOLID     GEOMETRY 

PROPOSITION  VIII.     THEOREM 

769.  The  intersection  of  two  spherical  surfaces  is  the 
circumference  of  a  circle  whose  plane  is  perpendicular  to  the 
line  joining  the  centers  of  the  spheres,  and  whose  center  is  w 
that  line. 


Given  two  intersecting  ©  0  and  0'  which,  by  rotation 
about  the  line  00'  as  an  axis,  generate  two  intersecting 
spherical  surfaces. 

To  prove  that  the  intersection  of  the  spherical  surfaces 
is  a  O,  whose  plane  J_  00',  and  whose  center  lies  in  00'. 

Proof.  Let  the  two  circles  intersect  in  the  points  P  and 
Q,  and  draw  the  common  chord  PQ, 

Then,  as  the  two  given  ©  rotate  about  00'  as  an  axis, 
the  point  P  will  generate  the  line  of  intersection  of  the  two, 
spherical  surfaces  that  are  formed. 

But  PR  is  constantly  J_  00'.  Art.  241, 

/.  PR  generates  a  plane  A.  00f  Art.  510. 

Also  PR  remains  constant  in  length. 

.*.  P  describes  a  circumference  in  that  plane.       Art.  197. 

Hence  the  intersection  of  two  spherical  surfaces  is  a  O , 
whose  plane  J_  the  line  of  centers,  and  whose  center  is  in 
the  line  of  centers. 

Q.  E.  D. 

The  above  demonstration  is  an  illustration  of  the  use  of  the  second 
definition  of  a  sphere  (Art.  732). 


THE     SPHERE  437 

PROPOSITION  IX.     THEOREM 

770.  A  spherical  angle  is  measured  by  the  arc  of  a  great 
circle  described  from  the  vertex  of  the  angle  as  a  pole,  and 
included  between  its  sides,  produced,  if  necessary. 


Given  /.BAG  a  spherical  angle  formed  by  the  intersec- 
tion of  the  arcs  of  the  great  circles  BA  and  CA,  and  BO 
an  arc  of  a  great  circle  whose  pole  is  A. 

To  prove  /.BAG  measured  by  arc  BC. 

Proof.  Draw  AD  tangent  to  AB,  and  AF  tangent  to 
AC.  Also  draw  the  radii  OB  and  OG. 

Then  AD  _L  AO.  Art.  230. 

Also  OB  _L  AO  ( for  AB  is  a  quadrant) . 

:.  OB  \\AD.  (WhyT) 

Similarly  OG\\AF. 

:.  Z  BOG=  Z  DAP  Art.  538. 

But  /.BOG  is  measured  by  arc  BG.  Art.  257. 

/.  /.DAF,  that  is,  /.BAG,  is  measured  by  arc  BG. 

Us-  (Why?) 

Q.  E.  D. 

771.  COR.  A  spherical  angle  is  equal  to  the  plane  angle 
of  the  dihedral  angle  formed  by  the  planes  of  its  sides. 


438  BOOK   IX.      SOLID     GEOMETRY 

SPHERICAL   TRIANGLES  AND   POLYGONS 

772.  A  spherical  polygon  is  a  portion  of  the  surface  of 
a  sphere  bounded  by  three  or  more 

arcs  of  great  circles,  as  ABCD. 

The  sides  of  the  spherical  polygon 
are  the  bounding  arcs;  the  vertices 
are  the  points  in  which  the  sides  in- 
tersect; the  angles  are  the  spherical 
angles  formed  by  the  sides. 

The  sides  of  a  spherical  polygon  are  usually  limited  to  arcs  less 
than  a  semicircumference. 

773.  A  spherical  triangle  is  a  spherical  polygon  of  three 
sides. 

Spherical  triangles  are  classified  in  the  same  way  as 
plane  triangles;  viz.,  as  isosceles,  equilateral,  scalene, 
right,  obtuse  and  acute. 

774.  Relation  of  spherical  polygons  to  polyhedral  angles. 
If  radii  be  drawn  from  the  center  of  a  sphere  to  the  ver- 
tices of  a  spherical   polygon  on  its  surface   (as   OA,    OB, 
etc.,  in  the  above  figure),  a  polyhedral  angle  is  formed  at 
O,  which  has  an  important  relation  to  the  spherical  poly- 
gon ABCD 

Each  face  angle  of  the  polyhedral  angle  equals  (in  num- 
ber of  degrees  contained)  the  corresponding  side  of  the  spheri- 
cal polygon; 

Each  dihedral  angle  of  the  polyhedral  angle  equals  the 
corresponding  angle  of  the  spherical  polygon. 

Hence,  corresponding  to  each  property  of  a  polyhedral 
angle,  there  exists  a  property  of  a  spherical  polygon,  and 
conversely. 


THE     SPHERE  439 

Hence,  also,  a  trihedral  angle  and  its  parts  correspond 
to  a  spherical  triangle  and  its  parts. 

Of  the  common  properties  of  a  polyhedral  angle  and  a  spherical 
polygon,  some  are  discovered  more  readily  from  the  one  figure  and 
some  from  the  other.  In  general,  the  spherical  polygon  is  simpler 
to  deal  with  than  a  polyhedral  angle.  For  instance,  if  a  trihedral 
angle  were  drawn  with  the  plane  angles  of  its  dihedral  angles,  nine 
lines  would  be  used,  forming  a  complicated  figure  in  solid  space; 
whereas,  the  same  magnitudes  are  represented  in  a  spherical  triangle 
by  three  lines  in  an  approximately  plane  figure. 

On  the  other  hand,  the  spherical  polygon,  because  of  its  lack  of 
detailed  parts,  is  often  not  so  suggestive  of  properties  as  the  poly- 
hedral angle. 


PROPOSITION  X.     THEOREM 

775.    The  sum  of  two  sides  of  a  spherical  triangle  is 
greater  than  the  third  side. 


Given  the  spherical  triangle  ABC,  of  which  no  side  is 
larger  than  AB. 

To  prove  AC+  BC  >  AB. 

Proof.   From  the  center  of  the  sphere,  0,  draw  the  radii 
OA,  OB,  OC. 

Then,  in  the  trihedral  angle  0-ABC, 

£AOC+£BOC  >  AOB.  Art.  582. 

/.  AC-}-BG  >  AB.  Art.  774. 

Q.  £.  D. 


440  BOOK  IX.      SOLID    GEOMETRY 

776.  COR.  1.  Any  side  of  a  spherical  triangle  is  greater 
than  the  difference  between  the  other  two  sides. 

111.  COR.  2.  The  shortest  path  between  two  points  on 
the  surface  of  a  sphere  is  the  arc  of  the  great  circle  joining 
those  points. 

For  any  other  path  between  the  two  points  may  be 
made  the  limit  of  a  series  of  arcs  of  great  circles  connect- 
ing successive  points  on  the  path,  and  the  sum  of  this 
series  of  arcs  of  great  circles  connecting  the  two  points  is 
greater  than  the  single  arc  of  a  great  circle  connecting 
them. 

PROPOSITION  XI.     THEOREM 

778.  The  sum  of  the  sides  of  a  spherical  polygon  is  less 
than  360°. 


Given  the  spherical  polygon  ABCD. 

To  prove  the  sum  of  the  sides  of  ABCD  <  360°. 

Proof.   From  0,  the  center  of  the  sphere,  draw  the  radii 
OA,  OB,  OC,  OD. 

Then  Z  AOB  +  Z.BOC  +  Z  COD-}-  ^DOA  <  300°. 

(Why  ?) 

/.  AB  +  BG  +  GD  +  DA  <  360° .  Art.  774. 

Q.  £.  D. 


SPHEKICAL     TKIANGLES 


441 


D 


A' 


779.    DEF.    The  polar  p'    D' 

triangle  of  a  given  triangle 
is  the  triangle  formed  by 
taking  the  vertices  of  the 
given  triangle  as  poles,  and 
describing  arcs  of  great  cir- 
cles. (Hence,  if  each  pole 
be  regarded  as  a  center,  the  radius  used  in  describing  each 
arc  is  a  quadrant.)  Thus  A'B'C 'is  the  polar  triangle  of 
ABC-,  also  D'E'F'  is  the  polar  triangle  of  DEF. 


PROPOSITION  XII.     THEOREM 

780.    //  one  spherical  triangle  is  the  polar  of  another, 
then  the  second  triangle  is  the  polar  of  the  first. 


Given  A'B'C'  the  polar  triangle  of  ABC. 

To  prove  ABC  the  polar  triangle  of  A'B'C'. 

Proof.   B  is  the  pole  of  the  arc  A'C'.  Art.  779. 

.*.  arc  A'B  is  a  quadrant.  (Why  ?) 

Also  C  is  the  pole  of  the  arc  A'B'.  (Why  ?) 

.*.  arc  A'C  is  a  quadrant.  (Why  ?) 

/.  A'  is  at  a  quadrant's  distance  from  both  B  and  C. 

/.  Af  is  the  pole  of  the  arc  BC.  Art.  750. 
In  like  manner  it  may  be  shown  that  B'  is  the  pole  of 

AC,  and  C'  the  pole  of  AB.  q.  E.  j>. 


442  BOOK   IX.       SOLID     GEOMETRY 

PROPOSITION  XIII.     THEOREM 

781.  In  a  spherical  triangle  and  its  polar,  each  angle  of 
one  triangle  is  the  supplement  of  the  side  opposite  in  the 
other  triangle. 

A' 


Given  the  polar  A  ABC  and  A'B'C'  with  the  sides  of 
ABC  denoted  by  a,  &,  c,  and  the  sides  of  A'B'C'  denoted 
by  a',  &',  c',  respectively. 

To  prove  A  +  a'=  180°,  B  +  &'  =  180°,  C  +  c'  =  180°f 
A'  +  a  =  180°,  JB'  +  6  =  180°,  (7  +  c  =  180°. 

Proof.  Produce  the  sides  AB  and  A G  till  they  meet 
B'O  in  the  points  D  and  .F,  respectively. 

Then  B'  is  the  pole  of  AF  :.  arc  5^=90°.         Art.  780. 

Also  C'  is  the  pole  of  AD  :.  arc  0/D  =  90°.          (Why?) 

Adding,  £'^  +  (71)=  180°.  (Why?) 

Or  J3'.F+  ^O7  +  D^-1800.  Ax.  6. 

Or  JB/C/  +  D^T-180°. 

But  B'C'  =  a',  and  D^is  the  measure  of  the  Z  A.  Art.  770. 
/.  A  +  a' =  180°. 

In  like  manner  the  other  supplemental  relations  may  be 
proved  as  specified. 

Q.  E.  D. 

782.  DEF.  Supplemental  triangles  are  two  spherical 
triangles  each  of  which  is  the  polar  triangle  of  the  other. 

This  new  name  for  two  polar  triangles  is  due  to  the 
property  proved  in  Art.  781. 


SPHERICAL    TRIANGLES  443 

PROPOSITION  XIV.     THEOREM 

783.    The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  180°,  and  less  than  540°. 


Given  the  spherical  triangle  ABO. 
To  prove       A  +  B  -f  G  >  180°  and  <  540°. 
Proof.     Draw   A'B'C',  the   polar   triangle   of   ABC,  and 
denote  its  sides  by  «',  b' ',  c' . 

Then  ^  +  a'  =  180°  ) 

J?  _j-  £'    —  180°  f  Art.  781. 

(7  _j_  c'  =  180°  ) 

.-.  A  +  J5  +  (7  +  «'  +  ^  +  c'  =  540°.     .     .     (1)       AX.  2. 
But  C  a'  +  £'  -f  cr  <  360°  Art.  778. 

{  a'  +  J'  +  cr  >  0° 
Subtracting  each  of  these  .in  turn  from  (1), 

^  _j_  B  +  C  >  180°  and  <  540°.  Ax.  11. 

Q.  E.  D. 

784.  COR.    A  spherical  triangle  may  have  one,  two  or 
three  right  angles;  or  it  may  have  one,  two  or  three  obtuse 
angles. 

785.  'DEF.    A  birectangular  spherical  triangle  is  a  spher- 
ical triangle  containing  two  right  angles. 

786.  DEF.    A   trirectangular    spherical    triangle    is    a 
spherical  triangle  containing  three  right  angles. 


444 


BOOK  IX.      SOLID     GEOMETRY 


787.  COR.    The  surface  of  a  sphere  may  be  divided  into 
eight  trirectangular  spherical  triangles.    For  let  three  planes 
J_  to  each  other  be  passed  through  the  center  of  a  sphere,  etc. 

788.  DBF.    The  spherical  excess  of  a  spherical  triangle 
is  the  excess  of  the  sum  of  its  angles  over  180°. 

789.  DEF.    Symmetrical  spherical  triangles  are  triangles 
which  have  their  parts  equal,  but  arranged  in  reverse  order. 


B' 


Fig.  2 


Fig-.  3 


Fig.l 


'  Three  planes  passing  through  the  center  of  a  sphere 
form  a  pair  of  symmetrical  spherical  triangles  on  opposite 
sides  of  the  sphere  (see  Art.  580),  as  A  ABC  and  A'B'C* 
of  Fig.  1. 

790.    Equivalence   of    symmetrical  spherical  triangles. 
Two  plane  triangles  which  have  their  parts  equal,  but  ar- 
ranged in  reverse 
order,    may    be 
made  to  coincide 
by  lifting  up  one 
triangle,  turning 
it  over  in  space,  and  placing  it  upon  the  other  triangle. 

But  two  symmetrical  spherical  triangles  cannot-be  made 
to  coincide  in  this  way,  because  of  the  curvature  of  a 
spherical  surface.  Hence  the  equivalence  of  two  sym- 
metrical spherical  triangles  must  be  demonstrated  in  some 
indirect  way. 


SPHE1UCAL    TRIANGLES 


445 


791.  Property  of  symmetrical  spherical  triangles.     Two 
isosceles  symmetrical  spherical  triangles  are  equal,  for  they 
can  be  made  to  coincide. 

PROPOSITION  XV.     THEOREM 

792.  Two  symmetrical  spherical  triangles  are  equivalent. 


Given  the  symmetrical  spherical  A  ABC  and  A'B'C', 
formed  by  planes  passing  through  0,  the  center  of  a  sphere. 
(See  Art.  789.) 

To  prove  A  ABC=a=  A  A'B'C'. 

Proof.  Let  Pbe  the  pole  of  a  small  circle  passing  through 
the  points  A,  J3,  C.  Draw  the  diameter  POP'. 

Also  draw  PA,  PB,  PC,  P'A',  P'B',  P'C',  all  arcs  of 
great  ©. 

Then  PA  =  PB  =  PC.  Art.  747. 

Also        P'A'  =  PA,    P'B'  =  PB,    P'C'  =  PC. 

Arts.  78,  215. 
;  .%  P'A'  =  P'B'  =P'C'.  Ax.  1. 

Hence  PAB  and  P'A'B'  are  symmetrical  isosceles  A. 


Similarly              A  PA  C=  A  P'A'G'.  Art.  791. 
And                       A  PBC=  A  P'E'G'} 
Adding          A  PAB+  A  PAC+  A  PEG 

^  A  P*A'B'+  A  PA'O+  A  P'B'  a.  Ax.  2. 

Or                         A  ABC-  A  A'B'C'.  Ax.  6. 

In  case  the  poles  P  and  P'  fall  outside  the  A  A.B<7  and 

A'B'C1  ',  let  the  puuil  suoply  the  demonstration.  Q.  E.  D. 


446  BOOK  IX.      SOLID    GEOMETRY 

PROPOSITION  XVI.     THEOREM 

793.    On  the  same  sphere,  or  on  equal  spheres,  two  tri- 
angles are  equal, 

I.  //  two  sides  and  the  included  angle  of  one  are  equal  to 
two  sides  and  the  included  angle  of  the  other;  or 

II.  If  two  angles  and  the  included  side  of  one  are  equal 
to  two  angles  and  the  included  side  of  the  other, 

the  corresponding  equal  parts  being  arranged  in  the  same 
order  in  each  case. 


I.  Given   the  spherical   A  ABC  and  DEF,   in  which 
AC=DF,  CB=FE,  and  Z<7=  /.F. 

To  prove  A  ABC=  A  DEF. 

Proof.     Let  the  pupil  supply  the  proof  (see  Book  I, 
Prop.  VI). 

II.  Given    the  spherical  A  ABC  and  DEF,  in  which 


To  prove  A  ABC=  A  DEF. 

Proof.  Let  the  pupil  supply  the  proof  (see  Book  I, 
Prop.  VII)  .  _ 

Ex.  1.  If  the  line  of  centers  of  two  spheres  is  10  in.,  and  the  radii 
are  12  in.  and  3  in.,  how  are  the  spheres  situated  with  reference  to 
each  other  ? 

Ex.  2.  The  tank  on  a  motor  car  is  a  cylinder  35  inches  long  and 
15  inches  in  diameter.  How  many  gallons  of  gasolene  will  it  hold  ? 

Ex.  3.  In  an  equilateral  cone,  find  the  ratio  of  the  lateral  area  to 
the  area  of  the  base. 


SPHERICAL     TRIANGLES  447 

PROPOSITION  XVII.      THEOREM 

794.    On  the  same  sphere,  or  on  equal  spheres,  two  tri- 
angles are  symmetrical  and  equivalent, 

I.  If  two  sides  and  the  included  angle  of  one  are  equal  to 
two  sides  and  the  included  angle  of  the  other;  or 

II.  //  two  angles  and  the  included  side  of  one  are  equal 
to  two  angles  and  the  included  side  of  the  other, 

the  corresponding  equal  parts  being  arranged  in  reverse 
order. 

A 


I.  Given  the  spherical  A  ABC  and  DEF,  in  which  AB 

=  DE,  AC  =  DF,  and  /.A  =  /.D,  the  corresponding  parts 
being  arranged  in  reverse  order. 

To  prove  A  ABC  symmetrical  with  A  DEF. 

Proof.    Construct  the  &D'E'F'  symmetrical  with  ADEF. 

>t 

Then  A  ABC  may  be  made  to  coincide  with  A  D'E'F', 

Art.  793. 

(having  two  sides  and  the  included  Z  equal  and  arranged  in  the 
same  order). 

But  A  D'E'F'  is  symmetrical  with  the  A  DEF. 

:.  A  ABC,  which  coincides  with  A  D'E'F',  is  symmet- 
rical with  A  DEF. 

II.  The  second  part  of  the  theorem  is  proved  in  the 

same  way. 

Q.  £.  D. 


448  BOOK   IX.      SOLID     GEOMETRY 

PROPOSITION  XVIII.     THEOREM 

795.  If  two  triangles  on  the  same  sphere,  or  equal 
spheres,  are  mutually  equilateral,  they  are  also  mutually 
equiangular,  and  therefore  equal  or  symmetrical. 


Given  two  mutually  equilateral  spherical  A  ABC  and 
A'B'C'  on  the  same  or  on  equal  spheres. 

To  prove  A  ABC  and  A'B'C'  equal  or  symmetrical. 

Proof.  From  0  and  0',  the  centers  of  the  spheres  to 
which  the  given  triangles  belong,  draw  the  radii  OA,  OB, 
OC,  O'A',  0'Bf,  O'C'. 

Then  the  face  A  at  0  =  corresponding  face  A  at  0' '. 

Art.  774. 

Hence  dihedral  A  at  0= corresponding  dihedral  A  at  Of. 

Art.  584. 

.*.  A  of  spherical  A  A BC=  homologous  A  of  spherical 

&  A'B'C'.  Art.  774. 

.*.  the  A  ABC  and  A'B'C'  are  equal  or  symmetrical, 
according  as  their  homologous  parts  are  arranged  in  the 
same  or  in  reverse  order.  Art.  789. 

796.  NOTE.  The  conditions  in  Props.  XVI  and  XVIII  which  make 
two  spherical  triangles  equal  are  the  same  as  those  which  make 
two  plane  triangles  equal.  Hence  many  other  propositions  occur  in 
spherical  geometry  which  are  identical  with  corresponding  proposi- 
tions in  plane  geometry.  Thus,  many  of  the  construction  problems  of 
spherical  geometry  are  solved  in  the  same  way  as  the  corresponding 
construction  problems  in  plane  geometry  j  as,  to  bisect  a  given 
angle,  etc, 


SPHERICAL    TRIANGLES  449 

PROPOSITION  XIX.     THEOREM 

797.  //  two  triangles  on  the  same  sphere  are  mutually 
equiangular,  they  are  also  mutually  equilateral,  and  there- 
fore equal  or  symmetrical. 


Given  the  mutually  equiangular  spherical  A  Q  and  Q* 
on  the  same  sphere  or  on  equal  spheres. 

To  prove  that  Q  and  Qf  are  mutually  equilateral,  and 
therefore  equal  or  symmetrical. 

Proof.  Construct  P  and  Pf  the  polar  A  of  Q  and  #', 
respectively. 

Then  A  P  and  Pf  are  mutually  equilateral.  Art.  781. 

/.  A  P  and  Pf  are  mutually  equiangular.  Art.  795. 

But  Q  is  the  polar  A  of  P,  and  Q'  of  P' '.  Art.  780. 

/.  A  Q  and  Q'  are  mutually  equilateral.  Art.  781. 

Hence  Q  and  Q'  are  equal  or  symmetrical,  according  as 
their  homologous  parts  are  arranged  in  the  same  or  in 
reverse  order.  Art.  789. 

Q.  E.  D. 

798.  COR.  If  two  mutually  equiangular  triangles  are 
on  unequal  spheres,  their  corresponding  sides  have  the  same 
ratio  as  the  radii  of  their  respective  spheres. 

CC 


450 


BOOK  IX.      SOLID     GEOMETRY 


PROPOSITION  XX.     THEOREM 

799.    In  an  isosceles  spherical  triangle  the  angles  oppo< 
site  the  equal  sides  are  equal. 


Given  the  spherical  A  ABC  in  which  AB=AC. 
To  prove  Z£=ZO. 

Proof.  Draw  an  arc  from  the  vertex  A  to  D,  the  mid- 
point of  the  base. 

Let  the  pupil  supply  the  remainder  of  the  proof. 

PROPOSITION  XXI.     THEOREM     (CONY.  OF  PROP.  XX) 

800.  If  two  angles  of  a  spherical  triangle  are  equal,  the 
sides  opposite  these  angles  are  equal,  and  the  triangle  is 
isosceles. 

A. 


Given  the  spherical  A  ABC  in  which  £B=  Z  G. 

To  prove  AB  =  AC. 

Proof.    Construct  A  A'B'C'  the  polar  A  of  ABC. 

Then  A'C'  =  A'B'.  Art.  781. 

/.    Z  C'=  Z-B'.  Art.  799. 

.'.  AB  =  AG.  Art.  781. 

Q.  £.  D. 


SPHEKICAL     TRIANGLES  451 

PROPOSITION  XXII.     THEOREM 

801.  In  any  spherical  triangle,  if  two  angles  are  un- 
equal, the  sides  opposite  these  angles  are  unequal,  and  the 
greater  side  is  opposite  the  greater  angle,  and  CONVERSELY^, 


Given  the  spherical  A  ABC  in  which  Z BAG  is  greater 
than  Z.C. 

To  prove  BC  >  BA. 

Proof.    Draw  the  arc  AD  making  /.D AC  equal  to  Z  C. 

Then                             DA  =  DC.  Art.  800. 
To  each  of  these  equals  add  the  arc  BD. 

:.  BD  +  DA  =  BD  +  DC,  or  EG.  (Why  f) 

But  in  A  BDA,  BD  +  DA  >  BA.  (Why  ?) 

/.  BC  >  BA.  .         Ax.  8. 

Let  the  pupil  prove  the  converse  by  the  indirect  method 
(see  Art.  106). 

Q.  E.  D. 

Ex.  1.    Bisect  a  given  spherical  angle. 
•    Ex.  2.    Bisect  a  given  arc  of  a  great  circle  on  a  sphere. 

Ex.  3.  At  a  given  point  in  an  arc  on  a  sphere,  construct  an  angle 
equal  to  a  given  spherical  angle  on  the  same  sphere. 

Ex.  4.  Find  the  locus  of  the  centers  of  the  circles  of  a  sphere 
formed  by  planes  perpendicular  to  a  given  diameter  of  the  given 
sphere. 


452 


BOOK   IX.       SOLID     GEOMETRY 


SPHERICAL    AREAS 

802.  Units   of   spherical  surface.     A  spherical  surface 
may  be  measured  in  terms  of,  either 

1.  The  customary  units  of  area,   as  a  square  inch,  a 
square  foot,  etc.,  or 

2.  Spherical  degrees,  or  spherids. 

803.  A   spherical   degree,  or  spherid,    is  one -ninetieth 
part  of  one  of  the  eight  trirectangular  triangles  into  which 
the  surface  of  a  sphere  may  be  divided  (Art.  787),  or  TTO 
part  of  the  surface  of  the  entire  sphere. 

A  solid  degree  is  one-ninetieth  part  of  a  trirectangular  angle  (see 
Art.  774). 

804.  A  lune  is  a  portion  of   the  surface  of  a  sphere 
bounded  by  two  semicircumferences 

of  great  circles,  as  PBP'C  of  Fig.  1, 
The  angle  of  a  lune  is  the  angle 
formed    by   the    semicircumferences 
which  bound  it,  as  the  angle  BPC. 

805.  A  zone  is  the  portion  of  the 
surface   of   the  sphere   bounded  by 
two  parallel  planes. 

A  zone  may  also  be  defined  as  the  sur- 
face generated  by  an  arc  of  a  revolving 
semicircumference.  Thus,  if  QFQ'  (Fig.  2) 
generates  a  sphere  by  rotating  about  QQf, 
its  diameter,  any  arc  of  QFQ',  as  EF, 
generates  a  zone. 

806.  A  zone  of  one  base  is  a  zone 
one   of   whose    bounding   planes   is 
tangent  to  the  sphere,   as  the  zone 
generated  by  the  arc  QE  of  Fig.  2. 

807.  The  altitude  of  a  zone  is  the  perpendicular  dis- 
tance between  the  bounding  planes  of  the  zone. 

The  bases  of  a  zone  are  the  circumferences  of  the  circles 
of  the  sphere  formed  by  the  bounding  planes  of  the  zone. 


SPHERICAL    AREAS 


453 


PROPOSITION  XXIII.     THEOREM 

0 

808.    The  area  generated   ~by  a  straight  line  revolving 

about  an  axis  in  its  plane  is  equal  to  the  projection  of  the 
line  upon  the  axis,  multiplied  by  the  circumference  of  a  circle 
ivhose  radius  is  the  perpendicular  erected  at  the  midpoint  of 
the  line  and  terminated  by  the  axis. 


Given  AB  and  XYin  the  same  plane,  CD  the  projection 
of  AB  on  JTY,  P#  the  -L  bisector  of  A£;  and  a  surface 
generated  by  the  revolution  of  AB  about  XT,  denoted  as 


To  prove          area  AB=CDX2  nPQ. 
Proof.    1.  In  general,  the  surface  generated  by  AB  is 
the  surface  of  a  frustum  of  a  cone  (Fig.  1). 
/.  area  AB  =  AB  X  2  nPR. 
Draw  AF  LED,  then 
A  ABF  and  PQR  are  similar. 
/.   A£:  AF=PQ:  PR. 
:.  AB  X  PR  =  AFX  PQ,  or  CD  X  PQ. 
Substituting,  area  AB=  CD  X  2  nPQ. 

2.  If  AB\\XY  (Fig.  2),  the  surface  generated  by  AB 
is  the  lateral  surface  of  a  cylinder. 

/.  area  AB=CDX2  nPQ.  Art.  697. 

3.  If  the  point  A  lies  in  the  axis  XY  (Fig.  3),  let  the 
pupil  show  that  the  same  result  is  obtained.  p.  E.  p, 


Art.  728. 

Art.  328. 

(Why?) 

(Why?) 

Ax.  8. 


454 


BOOK   IX.       SOLID     GEOMETKY 


PROPOSITION  XXIV.     THEOREM 

809.  The  area  of  the  surface  of  a  sphere  is  equal  to  the 
product  of  the  diameter  of  me  sphere  by  the  circumference 
of  a  great  circle. 

A 


Given  a  sphere  generated  by  the  revolution  of  the  semi- 
circle ACE  about  the  diameter  AE,  with  the  surface  of  the 
sphere  denoted  by  S,  and  its  radius  by  R. 
To  prove  8=AE  X  2  nR. 

Proof.    Inscribe  in  the  given  semicircle  the  half  of  a 
regular  polygon  of  an  even  number  of  sides,  as  ABCDE. 

Draw  the  apothem  to  each  side  of  the  semipolygon,  and 
denote  it  by  a. 

From  the  vertices  B,  C,  D  draw  Ja  to  AE. 
Then  area  AB  =  AF  X  2  na. 

area  BC=  FOX  2  na. 
area  CD=OKX2  na. 
area  DE=KEX  2  na. 
Adding,  area  ABCDE=AE  X  2  na. 
If,  now,  the  number  of  sides  of  the  polygon  be  indefi- 
nitely increased, 

area  ABCDE  approaches  8  as  a  limit.      Art.  441. 
And  a  approaches  R  as  a  limit.  (Why?) 

.*.  AE  X  2  na  approaches  AE  X  2  nR  as  a  limit,  f  Why  ?) 
J3ut         area  ABCDE '=AE  X  2  na  always. 

,-,  $=AEX2  nil.  (WhyT) 

.  E.  9. 


Art.  808. 


SPHERICAL    AREAS  455 

810.  Formulas  for  area  of  surface  of  a  sphere. 
Substituting  for  AE  its  equal  2  R,  #=4  nR2. 
Also  denoting  the  diameter  of  the  sphere  by  D,  R=  J  D. 


811.  Con.  1.  T/ie  surface  of  a  sphere  is  equivalent  to  four 
times  the  area  of  a  great  circle  of  the  sphere. 

812.  COR.  2.  The  areas  of  the  surfaces  of  tivo  spheres 
are  to  each  other  as  the  squares  of  their  radii,  or  of  their 
diameters. 

For,  if  S  and  $'  denote  the  surfaces,  R  and  R'  the  radii, 
and  D  and  Df  the  diameters  of  two  spheres, 


8  =  4KR2  =R2  8  =  KD2  =  D2 

S'    InR'2    R'2'       !°  8'    TiD'2    D'2' 

813.  Property  of  the  sphere.     The  following  property  of 
the  sphere  is  used  in  the  proof  of  Art.  809  :  //,  in  the  generat- 
ing arc  of  any  zone,  a  broken  line  be  inscribed,  whose  vertices 
divide  the  arc  into  equal  parts,  then,  as  the  number  of  these 
parts  is  increased  indefinitely,  the  area  generated  by  the  broken 
line  approaches  the  area  of  the  zone  as  a  limit.     Hence 

COR.  3.  The  area  of  a  zone  is  equal  to  the  circumference 
of  a  great  circle  multiplied  by  the  altitude  of  the  zone. 

Thus  the  area  generated  by  the  arc  BC  =  FOX2  rtR. 

814.  COR.  4.  On  the  same  sphere,  or  on  equal  spheres,  the 
areas  of  two  zones  are   to  each  other  as   the  altitudes  of  the 
zones.  _ 

Ex.  1.  Find  the  area  of  a  sphere  whose  diameter  is  10  in. 
Ex.  2.  Find  the  area  of  a  zone  of  altitude  3  in.,  on  a  sphere  whose 
radius  is  10  in. 


456 


BOOK  IX.      SOLID    GEOMETKY 


,      PROPOSITION  XXV.     THEOREM 

815.    The  area  of  a  lune  is  to  the  area  of  the   sur- 
face of  the  sphere  as  the   angle  of  the  lune  is  to  four  right 
is. 


Given  a  sphere  having  its  area  denoted  by  8,  and  on  the 
sphere  the  lune  ABCD  of  Z  A  with  its  area  denoted  by  L. 


To  prove 


L:S=A°  :  360°. 


Proof.  Draw  FBH,  the  great  O  whose  pole  is  A,  inter- 
secting the  bounding  arcs  of  the  lune  in  B  and  D. 

CASE  I.  When  the  arc  BD  and  the  circumference  FBH 
are  commensurable. 

Find  a  common  measure  of  BD  and  .FBH,  and  let  it  be 
contained  in  the  arc  BD  m  times,  and  in  the  circumference 
FBHn  times. 

Then      arc  BD  :  circumference  FBH=m  :  n. 

Through  the  diameter  AC,  and  the  points  of  division  of 
the  circumference  FBH  pass  planes  of  great  © . 

The  arcs  of  these  great  ©  will  divide  the  surface  of  the 
sphere  in  n  small  equal  lunes,  m  of  them  being  contained 
in  the  lune  ABCD. 

:.  L  :  S=m  :  n. 
:.  L  :  $=arc  BD  :  circumference  FBH.          (Why  ?) 

Or  L  :  S  =  A°  :  360°.  Art.  257. 

CASE  II.  When  the  arc  BD  and  the  circumference  FBH 
are  incommensurable. 

Let  the  pupil  supply  the  proof.  Q.  E.  D. 


SPHERICAL    AREAS  457 

816.  Formula  for  the  area  of  a  lune  in  spherical  degrees, 
or  spherids.     The  surface  of  a  sphere  contains  720  spherids 
(Art.  803).     Hence,  by  Art.  815, 

L       L  spherids        A        m   L_ 
''  fl°r720  spherids" 360'      *  2  " 
that  is,  the  area  of  a  lune  in  spherical  degrees  is  equal  to 
twice  the  number  of  angular  degrees  in  the  angle  of  the  lune. 

817.  Formula  for  area  of  a  lune  in  square  units  o 
S=4  TtR*  (Art.  810)   /.r^2-^,  or(L  =  ^g- 


818.  COE.  1.   On  the  same  sphere,  or  on  equal  spheres, 
two  lunes  are  to  each  other  as  their  angles. 

819.  COR.  2.   Two  lunes  ivith  equal  angles,  but,  on  un- 
equal spheres,  are  to  each  other  as  the  squares  of  the  radii  of 
their  spheres. 


'  Ex.  1.    Find  the  area  in  spherical  degrees  of  a  lune  of  27°. 

"Ex.  2.    Find  the  number  of  square  inches  in  the  area  of  a  lune  oft 
27°,  on  a  sphere  whose  radius  is  10  in. 

A  solid  symmetrical  with  respect  to  a  plane  is  a  solid  in  which 
a  line  drawn  from  any  point  in  its  surface  JL  the  given  plane  and 
produced  its  own  length  ends  in  a  point  on  the  surface ;  hence 

Ex.  3.  How  many  planes  of  symmetry  has  a  circular  cylinder  ?  A 
cylinder  of  revolution  ? 

Ex.  4.    Has  either  of  these  solids  a  center  of  symmetry  f 
Ex.  5.    Answer  the  same  questions  for  a  circular  cone. 
Ex.  6.    For  a  cone  of  revolution.     For  a  sphere. 

^     Ex.  7.    For  a  regular  square  pyramid.     For  a  regular  pentagonal 
pyramid. 


0 


458  BOOK  IX.      SOLID     GEOMETKY 

»  PROPOSITION  XXVI.     THEOREM 

820.  If  two  great  circles  intersect  on  a  hemisphere,  the 
sum  of  two  vertical  triangles  thus  formed  is  equivalent  to  a 
lune  ivhose  angle  is  that  angle  in  the  triangles  which  is 
formed  by  the  intersection  of  the  two  great  circles. 


Given  the  hemisphere  ADBF,  and  on  it  the  great  circles 
AFB  and  DFC,  intersecting  at  F. 

To  prove  A  AFC+  A  BFD  o  lune  whose  Z  is  BFD.  ' 

Proof.    Complete  the  sphere  and  produce  the  given  arcs 
of  the  great  circles  to  intersect  at  F'  on  the  other  hemisphere. 

Then,  in  the  A  AFG  and  BF'D, 

arc  AF=arc  BF' , 
(each  being  the  supplement  of  the  arc  BF). 

In  like  manner  arc  CF= 'arc  DF'. 
And  arc  AC  =  arc  DB. 

.  ••  A  AFC^  A  BF'D.  Art.  795. 

Add  the  A  BFD  to  each  of  these  equals ; 

/.  A  AFC  +  A  BFD<*  A  BF'D  +  A  BFD.      Ax.  3. 
Or    .-.  A  AFC  +  A  ^D^lune  FBF'D,  Ax.  6 

Q.  E.  D, 


SPHERICAL    AREAS  459 

PROPOSITION  XXVII.     THEOREM 

821.  The  number  of  spherical  degrees,  or  spherids,  in 
the  area  of  a  spherical  triangle  is  equal  to  the  number  of 
angular  degrees  in  the  spherical  excess  of  the  triangle. 


Given  the  spherical  A  ABO  whose  A  are  denoted  by  A , 
B,  C,  and  whose  spherical  excess  is  denoted  by  E. 

To  prove  area  of  A  ABC  —E  spherids. 

Proof.    Produce  the  sides  AC  and  BC  to  meet  AB  pro- 
duced in  the  points  D  and  F,  respectively. 

A  ABC  +  &  CDB  =  luuQ  ABDC=2A  spherids.  ) 

V      Art.  816. 

A  ABC  +  A  ACF=\unQ  BCFA  =  2  B  spherids.  j 

A  ABC  +  &  GFD=lune  of  /_BCA  =  2  C  spherids.     Art.  820- 

Adding,  and  observing  that  A  ABC+A  CDB+&ACF 
-f  A  CFD  =  hemisphere  ABDFC, 

2  A  ABC-}-  hemisphere  =  2  (A  +  B  +  C)  spherids.    Or, 
2  A  ABC+  360  spherids  =  2  (A  +  B  +  C)  spherids.    Ax.  8. 

/.  A  ABC  +  180  spherids  =  (A  +B'+  C)  spherids.  Ax.  5. 

/.  A  ABC=  ( A  +  B  +  C—180)  spherids.      Ax.  3. 

Or  area  A  ABC=E  spherids.  Art.  788. 

Q.  £.  D. 


460  BOOK  IX.      SOLID    GEOMETRY 

822,    Formula  for  area  of  a  spherical  triangle  in  square 
units  of  area. 

Comparing  the  area  of  a  spherical  A  with  the  area  of 
the  entire  sphere  . 

area  A  :  4  nR2  =  E  spherids  :  720  spherids. 


4  KR2  X  E  KR2E 

:.  area  A  =      -705—,  or  area  A--— 

823.    The  spherical  excess  of  a  spherical  polygon  is  the 

sum  of  the  angles  of  the  polygon  diminished  by  (n  —  2)  180°  ; 
that  is,  it  is  the  sum  of  the  spherical  excesses  of  the  tri- 
angles into  which  the  polygon  may  be  divided. 


PROPOSITION  XXVIII.     THEOREM 

824.  The  area  of  a  spherical  polygon,  in  spherical  de- 
grees or  spherids^  is  equal  to  the  spherical  excess  of  the 
polygon.  A 


Given  a  spherical  polygon  ABCDF  of  n  sides,  with  its 
spherical  excess  denoted  by  E. 

To  prove  area  of  ABCDF=E  spherical  degrees. 

Proof.  Draw  diagonals  from  A,  any  vertex  of  the  poly- 
gon, and  thus  divide  the  polygon  into  n — 2  spherical  A. 

The  area  of  each  A  =  (sum  of  its  A — 180)  spherids. 

Art.  821. 

/.  sum  of  the  areas  of  the  A  =  [sum  of  A  of  the  A — 
(n— 2)  180]  spherids.  Ax.  2. 

/.  area  of  polygon  =  E  spherids,  Art.  823. 

(for  the  sum  of  4  of  the  A— (n— 2)  180°=^°). 


SPHERICAL    VOLUMES 


461 


SPHERICAL   VOLUMES 

825.  A  spherical  pyramid  is  a  por- 
tion of  a  sphere  bounded  by  a  spheri- 
cal polygon  and   the   planes   of   the 
great  circles  forming  the  sides  of  the 
polygon.      The   base   of   a   spherical 
pyramid    is    the    spherical    polygon 
bounding  it,  and  the  vertex  of  the 
spherical  pyramid  is  the  center  of  the  sphere. 

Thus,  in  the  spherical  pyramid  O-ABCD,  the  base  is 
A  BCD  and  the  vertex  is  0. 

826.  A  spherical  wedge  (or  ungula)  is  the  portion  of  a 
sphere  bounded  by  a  lune  and  the  planes  of  the  sides  of 
the  lune. 

827.  A  spherical  sector  is  the  portion  of  a  sphere  gene- 
rated by  a  sector  of  that  semicircle  whose  rotation  generates 
the  given  sphere. 


828.  The  base  of  a  spherical  sector  is  the  zone  gene- 
rated by  the  revolution  of  the  arc  of  the  plane  sector  which 
generates  the  spherical  sector. 

Let  the  pupil  draw  a  spherical  sector  in  which  the  base 
is  a  zone  of  one  base. 


462  BOOK   IX.       SOLID     GEOMETRY 

829.  A   spherical   segment   is   a  portion  of   a   sphere 
included  between  two  parallel  planes. 

The  bases  of  a  spherical  segment  are  the  sections  of  the 
sphere  made  by  the  parallel  planes  which  bound  the  given 
segment ;  the  altitude  is  the  perpendicular  distance  between 
the  bases. 

830.  A  spherical  segment  of  one  base  is  a  spherical  seg- 
ment  one  of   whose   bounding   planes  is  tangent   to  the 
sphere. 

PROPOSITION  XXIX.     THEOREM 

831.  The  volume  of  a  sphere  is  equal  to  one -third  the 
product  of  the  area  of  its  surface  by  its  radius. 


Given  a  sphere  having  its  volume  denoted  by  V,  sur- 
face by  $,  and  radius  by  R. 

To  prove  V=%$XR. 

Proof.  Let  any  polyhedron  be  circumscribed  about  the 
sphere. 

Pass  a  plane  through  each  edge  of  the  polyhedron  and 
the  center  of  the  sphere. 

These  planes  will  divide  the  polyhedron  into  as  many 
pyramids  as  the  polyhedron  has  faces,  each  pyramid  hav- 
ing a  face  of  the  polyhedron  for  its  base,  the  center  of  the 


SPHERICAL    VOLUMES  4G3 

sphere  for  its  vertex,  and  the  radius  of  the  sphere  for  its 
altitude. 

.*.  volume  of  each  pyramid  =  J  base  X  R.       (Why  ?) 

.*.  volume  of  polyhedron  =  J  (surface  of  polyhedron)  XR. 

If  the  number  of  faces  of  the  polyhedron  be  increased 

indefinitely,  the  volume  of  the  polyhedron  approaches  the 

,  volume  of  the  sphere  as  a  limit,   and  the  surface  of  the 

polyhedron   approaches   the   surface   of   the   sphere   as   a 

limit. 

Hence  the  volume  of  the  polyhedron  and  J  (surface  of 
the  polyhedron  )X  R,  are  two  variables  always  equal. 

Hence  their  limits  are  equal, 

Or  V=bSXR.  (Why?) 

Q.  £.  D. 

832.  Formulas  for  volume  of  a  sphere.     Substituting 
8=4:  TtR2,  or  &=7tD2  in  the  result  of  Art.  831, 

also  F= 

D 

833.  COR.  1.    The  volumes  of  two  spheres  are  to  each 
other  as  the  cubes  of  their  radii,  or  as  the  cubes  of  their 
diameters. 


834.  COR.  2.    The   volume  of  a  spherical  pyramid  is 
equal  to  one  -third  the  product  of  its  base  by  the  radius  of 
the  sphere. 

835.  COR.  3.    The  volume  of  spherical  sector  is  equal  to 
one  -third  the  product  of  its  base  (the  bounding  zone)  by  the 
radius  of  the  sphere, 


464 


BOOK  IX.      SOLID     GEOMETRY 


836.    Formula  for  the  volume  of  a  spherical  sector.    De- 
noting the  altitude  of  the  sector  by  H  and  the  volume  by  V, 
F=4  (area  of  zone)  X  J2, 
=  j(2  nEH)  E.  Art.  813. 


PROPOSITION  XXX.     THEOREM 

837.  The  volume  of  a  spherical  segment  is  equal  to  one- 
half  the  product  of  its  altitude  by  the  sum  of  the  areas  of 
its  bases,  plus  the  volume  of  a  sphere  whose  diameter  is  the 
altitude  of  the  segment. 


Given  the  semicircle  ABC  A'  which  generates  a  sphere 
by  its  rotation  about  the  diameter  A  A ';  ED  and  CF  semi- 
chords  _L  AA',  and  denoted  by  r  and  r';  and  DF  denoted 
by  tf. 

To  prove  volume  of  spherical  segment  generated  by 
BCFD,  or  F=J(7ir2  +  7ir/2)  H+i  nH8. 

Proof.   Draw  the  radii  OB  and  0(7. 

Denote  OF  by  h,  and  OD  by  k. 

Then     F=vol.  OBC  +  vol.  OCF—  vol.  OBD. 

.'.    F=§  7l#2#-f  4  7tr'2h— J  7tr2A'.       Arts.  836,  723. 

But        H=h-k,  h*=lP-r*9  and  /t2=^2-r2,         (Why?) 


SPHERICAL    VOLUMES  465 

.-.  F=4  n  [2  R2  (h-k)-i  (R2-h2)  h—(R2-k2)  k] .    Ax.  8. 
=  4  7t  [2  R2'(h-k)  +  R2 'U-AO  — (A3-*3)]. 
=  i  7t  JT  [3  £2— U2  +  M  +  k2)]  .  x 

But  7i2— 2  M  -{-  k2  =  H2.  s  As.  4. 

,<* 

Subtract  each  member  from  3  ft2  -f  3  &2  and  divide  by  2. 

£*s.  3,  5. 

Then  A2  +  M  +  ^2  =  l  (/i2  +  k2)—- 


/.  F=  i  7t#  [f  (r2  +  r'2)  +  Y  ]  •  Ax.  8. 

Q.  E.  D. 

838.  Formula  for  volume  of  a  spherical  segment  of  one 
base.  In  a  spherical  segment  of  one  base  rf  =  o,  and  r2= 
(2  R-H)  H  (Art.  343). 

Substituting  for  r  and  r'  these  values  in  the  result  of 
Art.  837, 


839.  Advantage  of  measurement  formulas.  The  student 
should  observe  carefully  that,  by  the  results  obtained  in 
Book  IX,  the  measurement  of  the  areas  of  certain  curved 
surfaces  is  reduced  to  the  far  simpler  work  of  the  measure- 
ment of  the  lengths  of  one  or  more  straight  lines;  in  like 
manner  the  measurement  of  certain  volumes  bounded  by  a 
curved  surface  is  reduced  to  the  simpler  work  of  linear 
measurements.  A  similar  remark  applies  to  the  results  of 
Book  VIII.  

Ex.  1.    Find  the  volume  of  a  sphere  whose  radius  is  7  in. 
Ex.  2.    Find  the  volume  of  a  sphere  whose  diameter  is  7  in. 
Ex.  3.    In  a  sphere  whose  radius  is  8  in.,  find  the  volume  of  a 
spherical  segment  of  one  base  whose  altitude  is  3. 

DD 


466  BOOK   IX.       SOLID     GEOMETRY 

EXERCISES.     GROUP. 72 

THEOREMS    CONCERNING    THE    SPHERE 

Ex.  1.  Of  circles  of  a  sphere  whose  planes  pass  through  a  given 
point  within  a  sphere,  the  smallest  is  that  circle  whose  plaiie  is 
perpendicular  to  the  diameter  through  the  given  point. 

Ex.  2.  If  a  point  on  the  surface  of  a  given  sphere  is  equidistant 
from  three  points  on  a  given  small  circle  of  the  sphere,  it  is  the  pole 
of  the  small  circle. 

Ex.  3.  If  two  sides  of  a  spherical  triangle  are  quadrants,  the  third 
side  measures  the  angle  opposite  that  side  in  the  triangle. 

Ex.  4.  If  a  spherical  triangle  has  one  right  angle,  the  sum  of  its 
other  two  angles  is  greater  than  one  right  angle. 

Ex.  5.  If  a  spherical  triangle  is  isosceles,  its  polar  triangle  is 
isosceles. 

Ex.  6.  The  polar  triangle  of  a  birectangular  triangle  is  birectan- 
gular. 

Ex.  7.  The  polar  triangle  of  a  trirectangular  triangle  is  identical 
with  the  original  triangle. 

Ex.  8.  Prove  that  the  sum  of  the  angles  of  a  spherical  quadri- 
lateral is  greater  than  4  right  angles,  and  less  than  8  right  angles. 
What,  also,  are  the  limits  of  the  sum  of  the  angles  of  a  spherical 
hexagon  ?  Of  the  sum  of  the  angles  of  a  spherical  w-gon  ? 

Ex.  9.  On  the  same  sphere,  or  on  equal  spheres,  two  birectan- 
gular triangles  are  equal  if  their  oblique  angles  are  equal. 

Ex.  10.  Two  zones  on  the  same  sphere,  or  on  equal  spheres,  are 
to  each  other  as  their  altitudes. 

Ex.  11.  If  one  of  the  legs  of  a  right  spherical  triangle  is  greater^ 
than  a  quadrant,  another  side  is  also  greater  than  a  quadrant. 

[SuG.  Of  the  leg  which  is  greater  than  a  quadrant,  take  the  end 
remote  from  the  right  angle  as  a  pole,  and  describe  an  arc.] 

Ex.  12.  If  ABC  and  A'B'C'  are  polar  triangles,  the  radius  OA  is 
perpendicular  to  the  plane  OB'C'. 


EXEKCISES     ON     THE     SPHEKE  467 

Ex.  13.  On  the  same  sphere,  or  on  equal  spheres,  spherical  tri- 
angles whose  polar  triangles  have  equal  perimeters  are  equivalent. 

Ex.  14.  Given  OAO' ',  OBO> ',  and  AB  arcs  of  great 
circles,  intersecting  so  that  Z  OAB=  Z  CfBA;  prove  that 
AOA£=A(yAB.  A 

[Sue.   Show  that  /.OBA  =  /.WAS.] 

Ex.  15.  Find  the  ratio  of  the  volume  of  a  sphere  to 
the  volume  of  a  circumscribed  cube. 

Ex.  16.    Find  the  ratio  of  the  surface  of  a  sphere  to 
the  lateral    surface  of  a  circumscribed  cylinder  of  revolution;  also 
find  the  ratio  of  their  volumes. 

Ex.  17.  If  the  edge  of  a  regular  tetrahedron  is  denoted  by  a,  find 
the  ratio  of  the  volumes  of  the  inscribed  and  circumscribed  sphez'js. 

Ex.  18.  Find  the  ratio  of  the  two  segments  into  which  a  hemi- 
sphere is  divided  by  a  plane  parallel  to  the  base  of  the  hemisphere  and 
at  the  distance  $ R  from  the  base. 


EXERCISES.     GROUP  73 

SPHERICAL    LOCI 

Ex.  1.  Find  the  locus  of  a  point  at  a  given  distance  a  from  the 
surface  of  a  given  sphere. 

Ex.  2.  Find  the  locus  of  a  point  on  the  surface  of  a  sphere  that;  is 
equidistant  from  two  given  points  on  the  surface. 

Ex.  3.  If,  through  a  given  point  outside  a  given  sphere,  tangent 
planes  to  the  sphere  are  passed,  find  the  locus  of  the  points  of 
tangency. 

Ex.  4.  If  straight  lines  be  passed  through  a  given  fixed  point  in 
space,  and  through  another  given  point  other  straight  lines  be  passed 
perpendicular  to  the  first  set,  njicl  the  locus  of  the  feet  of  the 
perpendiculars. 


468  BOOK   IX.       SOLID     GEOMETRY 

EXERCISES.     GROUP  74 

PROBLEMS    CONCERNING    THE    SPHERE 

Ex.  1.    At  a  given  point  on  a  sphere,  construct  a  plane  tangent  to 
the  sphere. 

Ex.  2.    Through  a  given  point  on  the  surface  of  a  sphere,  draw  an 
arc  of  a  great  circle  perpendicular  to  a  given  arc. 

Ex.  3.    Inscribe  a  circle  in  a  given  spherical  triangle. 

Ex.  4.    Construct    a    spherical  triangle,    given  its  polar  triangle. 

Given  the   radius,  r,  construct  a  spherical   surface  which  shall  pass 

through 
Ex.  5.  -  Three  given  points. 

Ex.  6.  Two  given  points  and  be  tangent  to  a  given  plane. 

Ex.  7.  Two  given  points  and  be  tangent  to  a  given  sphere 

Ex.  8.  One  given  point  and  be  tangent  to  two  given  planes. 

Ex.  9.  One  given  point  and  be  tangent  to  two  given  spheres. 

Given   the   radius,  r,  construct  a  spherical  surface  which  shall   be 

tangent  to 
Ex.  10.    Three  given  planes. 

Ex.  11.    Two  given  planes  and  one  given  sphere. 

Ex.  12.  Construet  a  spherical  surface  which  shall  pass  through 
three  given  points  and  be  tangent  to  a  given  plane. 

Ex.  13.  Through  a  given  straight  line  pass  a  plane  tangent  to  a 
given  sphere. 

[SuG.  Through  the  center  of  the  sphere  pass  a  plane  X  given 
line,  etc.] 

When  is  the  solution  impossible  ? 

Ex.  14.  Through  a  given  point  on  a  sphere,  construct  an  arc  of 
a  great  circle  tangent  to  a  given  small  circle  of  the  sphere. 

[Sua.  Draw  a  straight  line  from  the  center  of  the  sphere  to  the 
given  point,  and  produce  it  to  intei-sect  the  plane  of  the  small  circle, 


NUMERICAL   EXERCISES   IN   SOLID 
GEOMETRY 

For  methods   of   facilitating   numerical   computations, 
see  Arts.  493-6. 

EXERCISES.     CROUP  75 

LINES    AND    SURFACES    OF    POLYHEDRONS 
Find  the  lateral  area  and  total  area  of  a  right  prism  whose 
/    Ex.  1.    Base  is  an  equilateral  triangle  of  edge  4  in.,  and  whose 
altitude  is  15  in. 

Ex.  2.    Base  is   a  triangle  of  sides  17,  12,  25,  and  whose  altitude 
is  20. 

f     Ex.  3.    Base  is  an  isosceles  trapezoid,  the  parallel  bases  being  10 
and  15  and  leg  8,  and  whose  altitude  is  24. 

*     Ex.  4.    Base    is  a  rhombus  whose  diagonals  are   12   and   16,  and 

whose  altitude  is  12. 

Ex.  5.    Base  is  a  regular  hexagon  with  side  8  ft.,  and  whose  alti- 
tude is  20  ft. 

Ex.  6.    Find   the  entire  surface   of  a    rectangular   parallelepiped 
8  X  12  X  16  in. ;  of  one  p  X  q  X  r  ft. 

Ex.  7.    Of  a  cube  whose  edge  is  1  ft.  3  in. 

Ex.  8.    The  lateral  area  of  a  regular  hexagonal  prism  is  120  sq.  ft. 
and  an  edge  of  the  base  is  10  ft.    Find  the  altitude. 

Ex.  9.    How  many  square  feet  of  tin  are  necessary  to  line  a  box 
20X6X4  in.  ? 

/  Ex.  10.    If  the  surface  of  a  cube  is  1  sq.  yd.,  find  an  edge  in  inches, 
Ex.  11.    Find  the  diagonal  of  a  cube  whose  edge  is  5  in. 
Ex.  13.    If  the  diagonal  of  a  cube  is  12  ft.,  find  the  surface, 


470  SOLID     GEOMETKY 

Ex.  13.  If  the  surface  of  a  rectangular  parallelepiped  is  208  sq.  in.," 
and  the  edges  are  as  2  :  3  : 4,  find  the  edges. 

In  a  regular  square  pyramid 

Ex.  14.  If  an  edge  of  the  base  is  16  and  slant  height  is  17,  find 
the  altitude. 

j    Ex.  15.    If  the  altitude  is  15  and  a  lateral  edge  is  17,  find  an  edge 
of  the  base. 

Ex.  16.  If  a  lateral  edge  is  25  and  an  edge  of  the  base  is  14,  find 
the  altitude. 

In  a  regular  triangular  pyramid 

Ex.  17.  If  an  edge  of  the  base  is  8  and  the  altitude  is  10,  find  the 
slant  height. 

Ex.  18.    Find  the  altitude  of  a  regular  tetrahedron  whose  edge  is  6. 

Find  the  lateral  surface  and  the  total  surface  of 

Ex.  19.  A  regular  square  pyramid  an  edge  of  whose  base  is  16, 
and  whose  altitude  is  15. 

Ex.  20.  A  regular  triangular  pyramid  an  edge  of  whose  base  is  10, 
and  whose  altitude  is  12. 

^     Ex.  21.    A  regular  hexagonal  pyramid  an  edge  of  whose  base  is  4, 
and  whose  altitude  is  21. 

Ex.  22.  A  regular  square  pyramid  whose  slant  height  is  24,  and 
•whose  lateral  edge  is  25. 

/     Ex.  23.    A  regular  tetrahedron  whose  edge  is  4. 
Ex.  24«.    A  regular  tetrahedron  whose  altitude  is  9. 

Ex.  25.  A  regular  hexagonal  pyramid  each  edge  of  whose  base  is 
a,  and  whose  altitude  is  6. 

Ex.  26.  In  the  frustum  of  a  regular  square  pyramid  the  edges  of 
the  bases  are  6  and  18,  and  the  altitude  is  8.  Find  the  slant  height. 
Hence  find  the  lateral  area. 

Ex.  27.  In  the  frustum  of  a  regular  triangular  pyramid  the  edges 
of  the  bases  are  4  and  6,  and  the  altitudi  is  5,  Find  the  slant  height. 
Hence  and  the  lateral  area. 


NUMERICAL    EXERCISES    IN    SOLID    GEOMETRY     471 

Ex.  28.  In  the  frustum  of  a  regular  tetrahedron,  if  the  edge  of  the 
lower  base  is  &i,  the  edge  of  the  upper  base  is  62,  and  the  altitude  is  a, 
show  that  £=iT/H&i  —  W  -f-4a2. 

Ex.  29.  In  the  frustum  of  a  regular  square  pyramid  the  edges  of 
the  bases  are  20  and  60,  and  a  lateral  edge  is  101.  Find  the  lateral 
surface. 

EXERCISES.     GROUP  76 

LINES    AND    SURFACES    OP    CONES    AND    CYLINDERS 

Ex.  1 .  How  many  square  feet  of  lateral  surface  has  a  tunnel  100 
yds.  long  and  7  ft.  in  diameter. 

Ex.  2.  The  lateral  area  of  a  cylinder  of  revolution  is  1  sq.  yd., 
and  the  altitude  is  1  ft.  Find  the  radius  of  the  base: 

Ex.  3.  The  entire  surface  of  a  cylinder  of  revolution  is  900  sq.  ft. 
and  the  radius  of  the  base  is  10  ft.  Find  the  altitude. 

In  a  cylinder  of  revolution 

Ex.  4.    Find  R  in  terms  of  8  and  H. 

Ex.  5.    Find  H  in  terms  of  E  and  T. 
Ex.  6.    Find  T  in  terms  of  S  and  H. 

Ex.  7.  How  many  sq.  yds-  of  canvas  are  required  to  make  a  coni- 
cal tent  20  ft.  in  diameter  and  12  ft.  high  ? 

Jt  Ex.  8.  A  man  has  400  sq.  yds.  of  canvas  and  wants  to  make  a 
conical  tent  20  yds.  in  diameter.  What  will  be  its  altitude  ? 

^  Ex.  9.  The  altitude  of  a  cone  of  revolution  is  10  ft.  and  the  lat- 
eral area  is  11  times  the  area  of  the  base.  Find  the  radius  of  the  base. 

In  a  cone  of  revolution 

Ex.  10.   Find  T  in  terms  of  S  and  L. 
Ex.  11.    Find  12  in  terms  of  T  and  L. 

J*  Ex.  12.  How  many  square  feet  of  tin  are  necessary  to  make  a 
funnel  the  diameters  of  whose  ends  are  2  in.  and  8  in.,  and  whose 
altitude  is  7  in.  ? 

It  Ex.  13.  If  the  slant  height  of  a  frustum  of  a  cone  of  revolution 
makes  an  angle  of  45°  with  the  base,  show  that  the  lateral  area  of  the 
frustum  is  (n2 — r22)  TT  i/2. 


472  SOLID    GEOMETRY 


EXERCISES.     CROUP  77 

SPHERICAL    LINES    AND    SURFACES 

Ex.  1.    Find   in  square  feet  the  area  of  the  surface  of  a  sphere 
whose  radius  is  1  ft.  2  in. 

f>  Ex.  2.  How  many  square  inches  of  leather  will  it  take  to  cover  a 
baseball  whose  diameter  is  3i  in.? 

Ex.  3.    How  many  sq.  ft.  of  tin  are  required  to  cover  a  dome  in 
the  shape  of  a  hemisphere  6  yds.  in  diameter  ? 

K  Ex.  4.    What  is  the  radius  of  a  sphere  whose  surface  is  616  sq.  in.  ? 
Ex.  5.    Find  the  diameter  of  a  globe  whose  surface  is  1  sq.  yd. 

Ex^jS.    If  the  circumference  of  a  great  circle  on  a  sphere  is  1  ft., 
find  the  area  of  the  surface  of  the  sphere. 

Ex.  7.    If  a  hemispherical  dome  is  to  contain  100  sq.  yds.  of  sur- 
face, what  must  its  diameter  be  ? 

*  Ex.  8.  Find  the  radius  of  a  sphere  in*  which  the  area  of  the  sur- 
face equals  the  number  of  linear  units  in  the  circumference  of  a  great 
circle. 

Find  the  area  of  a  lune  in  which 

X  Ex.  9.  The  angle  of  the  lune  is  36°,  and  the  radius  of  the  sphere 
is  14  in. 

Ex.  10.    The  angle  of  the  lune  is  18°  20',  and  the  diameter  of  the 
sphere  is  20  in. 

ft  Ex.  11.  The  angle  of  the  lune  is  24°,  and  the  surface  of  the 
sphere  is  4  sq.  ft. 

Find  the  area  of  a  spherical  triangle  in  which 

*  Ex.  12.  The  angles  are  80°,  90°,  120°,  and  the  diameter  of  the 
sphere  is  14  ft. 

+  Ex.  13.  The  angles  are  74°  24',  83°  16',  92°  20',  and  the  radius  of 
the  sphere  is  10. 

%  Ex.  14.  The  angles  are  85°,  95°,  135°,  and  the  surface  of  the 
sphere  is  10  sq.  ft. 


NUMERICAL    EXERCISES    IN    SOLID    GEOMETRY     473 

Ex.  15.  If  the  sides  of  a  spherical  triangle  are  100°,  110°,  120° 
and  the  radius  of  the  sphere  is  16,  find  the  area  of  the  polar  triangle. 

Ex.  16.  If  the  angles  of  a  spherical  triangle  are  90°,  100°,  120* 
and  its  area  is  3900,  find  the  radius  of  the  sphere. 

Ex.  17.  If  the  area  of  an  equilateral  spherical  triangle  is  one- 
third  the  surface  of  the  sphere,  find  an  angle  of  the  triangle. 

Ex.  18.  In  a  trihedral  angle  the  plane  angles  of  the  dihedral 
angles  are  80°,  90°,  100°;  find  the  number  of  solid  degrees  in  the 
trihedral  angle. 

S        Ex.  19.    Find   the  area  of    a  spherical   hexagon   each   of   whose 
angles  is  150°,  on  a  sphere  whose  radius  is  20  in. 

Ex.  20.  If  each  dihedral  angle  of  a  given  pentahedral  angle  is 
120°,  how  many  solid  degrees  does  the  pentahedral  angle  contain  ? 

Ex.  21.  In  a  sphere  whose  radius  is  14  in.,  find  the  area  of  a  zone 
3  in.  high. 

Ex.  22.  What  is  the  area  of  the  north  temperate  zone,  if  the 
earth  is  taken  to  be  a  sphere  with  a  radius  of  4,000  miles,  and  the 
distance  between  the  plane  of  the  arctic  circle  and  that  of  the  tropic 
of  Cancer  is  1,800  miles  ? 

Ex.  23.  If  Cairo,  Egypt,  is  in  latitude  30°,  show  that  its  parallel 
of  latitude  bisects  the  surface  of  the  northern  hemisphere. 

Ex.  24.  How  high  must  a  person  be  above  the  earth's  surface  to 
see  one  -third  of  the  surface  ? 


S/ 


Ex.  25.  How  much  of  the  earth's  surface  will  a  man  see  who  is 
2,000  miles  above  the  surface,  if  the  diameter  is  taken  as  8,000  miles? 

Ex.  26.  If  the  area  of  a  zone  equals  the  area  of  a  great  circle, 
find  the  altitude  of  the  zone  in  terms  of  the  radius  of  the  sphere. 

Ex.  27.  If  sounds  from  the  Krakatoa  explosion  were  heard  at  a 
distance  of  3,000  miles  (taken  as  a  chord)  on  the  surface  of  the  earth, 
over  what  fraction  of  the  earth's  surface  were  they  heard  ? 

Ex.  28.    The  radii  of  two  spheres  are  5  and  12  in.  and  their  cen- 
s  are  13  in.  apart.    Find  the  area  of  the  circle  of  intersection  and 

also  of  that  part  of  the  surface  of  each  sphere  not  included  by  the 

other  sphere. 


474  SOLID     GEOMETKY 

EXERCISES.     CROUP  78 

VOLUMES    OF    POLYHEDRONS 
Find  the  volume  of  a  prism 

XEx.  1.    Whose  base  is  an  equilateral  triangle  with  side  5  in.,  and 
rhose  altitude  is  16  in. 

)/     Ex.  2.    Whose  base  is  a  triangle  with  sides  12,  13,  15,  and  whose 
Altitude  is  20. 

Ex.  3.    Whose  base  is  an  isosceles  right  triangle  with  a  leg  equal 
to  2  yds.,  and  whose  altitude  is  25  ft. 

Ex.  4.   Whose  base  is  a  regular  hexagon  with  a  side  of  8  ft.,  and 
hose  altitude  is  10  yds. 

S    Ex.  5.    Whose  base  is  a  rhombus  one  of  whose  sides  is  25,  and  one 
of  whose  diagonals  is  14,  and  whose  altitude  is  11. 

Ex.  6.    Whose  base  contains  84  sq.  yds.,  and  whose  lateral  faces 
are  three  rectangles  with  areas  of  100,  170,  210  sq.  yds.,  respectively. 

XEx.  7.    How  many  bushels  of  wheat  are  held  by  a  bin  30  x  10  x  6  ft. , 
if  a  bushel  is  taken  as  li  cu.  ft.? 

Ex.  8.    How  many  cart-loads  of  earth  are  in  a  cellar  30  x 20x6  ft., 
if  a  cart-load  is  a  cubic  yard  ? 

/     Ex.  9.  'If  a  cubical  block  of  marble  costs  $2,  what  is  the  cost  of 
a  cube  whose  edge  is  a  diagonal  of  the  first  block  ? 

Ex.  10.    Find  the  edge  of  a  cube  whose  volume  equals  the  sum  of 
the  volumes  of  two  cubes  whose  edges  are  3  and  5  ft. 

Ex.  11.    Find  the  edge  of  a  cube  whose  volume  equals  the  area  of 
its  surface. 


» 


Ex.  12.    If  the  top  of  a  cistern  is  a  rectangle  12x8  ft.,  how  deep 
ust  the  cistern  be  to  hold  10,000  gallons  ? 


Ex.  13.  Find  the  inner  edge  of  a  peck  measure  which  is  in  the 
shape  of  a  cube. 

Ex.  14.  A  peck  measure  is  to  be  a  rectangular  parallelepiped  with 
square  base  and  altitude  equal  to  twice  the  edge  of  the  base,  Find 
its  dimensions, 


NUMERICAL    EXERCISES    IN    SOLID    GEOMETRY      475 

Ex.  15.    Find  the  volume  of  a  cube  whose  diagonal  is  a. 

Find  the  volume  of  a  pyramid 

Ex.  16.  Whose  base  is  an  equilateral  triangle  with  side  8  in.r 
and  whose  altitude  is  12  in. 

Ex.  17.  Whose  base  is  a  right  triangle  with  hypotenuse  29  and 
one  leg  21,  and  whose  altitude  is  20. 

Ex.  18.  Whose  base  is  a  square  with  side  6,  and  each  of  whose 
lateral  edges  is  5. 

Ex.  19.  Whose  base  is  a  square  with  side  10,  and  each  of  whose 
lateral  faces  makes  an  angle  of  45°  with  the  base. 

Ex.  20.  If  the  pyramid  of  Memphis  has  an  altitude  of  146  yds. 
and  a  square  base  of  side  232  yds.,  how  many  cubic  yards  of  stone 
does  it  contain  ?  What  is  this  worth  at  $1  a  cu.  yd.? 

Ex.  21.  A  church  spire  150  ft.  high  is  hexagonal  in  shape  and 
each  side  of  the  base  is  10  ft.  The  spire  has  a  hollow  hexagonal 
interior,  each  side  of  whose  base  is  6  ft.,  and  whose  altitude  is  45  ft. 
How  many  cubic  yards  of  stone  does  the  spire  contain  ? 

Ex.  22.  If  a  pyramid  contains  4  cu.  yds.  and  its  base  is  a  square 
with  one  side  2  ft.,  find  the  altitude. 

Ex.  23.  A  heap  of  candy  in  the  shape  of  a  frustum  of  a  regular 
square  pyramid  has  the  edges  of  its  bases  25  and  9  in.  and  its  altitude 
12  in.  Find  the  number  of  pounds  in  the  heap  if  a  pound  is  a  rectan- 
gular parallelepiped  4x3x2  in.  in  size. 

Ex.  24.  Find  the  volume  of  a  frustum  of  a  regular  triangular 
pyramid,  the  edges  of  the  bases  being  2  and  8,  and  the  slant  height  12. 

Ex.  25.  The  edges  of  the  bases  of  the  frustum  of  a  regular  square 
pyramid  are  24  and  6,  and  each  lateral  edge  is  15;  find  the  volume. 

Ex.  26.  If  a  stick  of  timber  is  in  the  shape  of  a  frustum  of  a 
regular  square  pyramid  with  the  edges  of  its  ends  9  and  15  in.,  and 
with  a  length  of  14  ft.,  find  the  number  of  feet  of  lumber  in  the  stick. 

What  is  the  difference  between  this  volume  and  that  of  a  stick  of 
the  same  length  having  the  shape  of  a  prism  with  a  base  equal  to  the 
area  of  a  midsection  of  the  first  stick  ? 


476  SOLID     GEOMETRY 

Ex.  27.    Find  the  volume  of  a  prismoid  whose  bases  are  rectangles 

5  x  2  ft.  and  7  x  4  ft.,  and  whose  altitude  is  12  ft. 

Ex.  28.  How  many  cart-loads  of  earth  are  there  in  a  railroad  cut 
12  ft.  deep,  whose  base  is  a  rectangle  100x8  ft.,  and  whose  top  is  a 
rectangle  30x50  ft.? 

Ex.  29.  Find  the  volume  of  a  prismatoid  whose  base  is  an  equi- 
lateral triangle  with  side  12  ft.,  and  whose  top  is  a  line  12  ft.  long 
parallel  to  one  side  of  the  base,  and  whose  altitude  is  1&  ft. 

Ex.  30.  If  the  base  of  a  prismatoid  is  a  rectangle  with  dimensions 
a  and  &,  the  top  is  a  line  c  parallel  to  the  side  b  of  the  base,  and  the 
altitude  is  h,  find  the  volume. 

EXERCISES.     CROUP  79 

VOLUMES    OF    CONES    AND    CYLINDERS 

Ex.  1.  How  many  barrels  of  oil  are  contained  in  a  cylindrical 
tank  20  ft.  long  and  6  ft.  in  diameter,  if  a  barrel  contains  4  cu.  ft.  f 

)^  Ex.  2.  How  many  cu.  yds.  of  earth  must  be  removed  in  making  a 
tunnel  450  ft.  long,  if  a  cross-section  of  the  tunnel  is  a  semicircle  of 
15  ft.  radius  ? 

V  Ex.  3.  A  cylindrical  glass  3  in.  in  diameter  holds  half  a  pint. 
Find  its  height  in  inches. 

y        Ex.  4.    If  a  cubic  foot  of  brass  be  drawn  out  into  wire  ^  inch  in 
^  diameter,  how  long  will  the  wire  be  ? 

Ex.  5.  A  gallon  measure  is  a  cylinder  whose  altitude  equals  the 
diameter  of  the  base.  Find  the  altitude.  * 

Ex.  6.  Show  that  the  volumes  of  two  cylinders,  having  the  altitude 
of  each  equal  to  the  radius  of  the  other,  are  to  each  other  as  E  :  E' . 

Ex.  7.  In  a  cylinder,  find  E  in  terms  of  V  and  H ;  also  V  in  terms 
of  8  and  E. 

A       Ex.  8.    A  conical  heap  of  potatoes  is  44  ft.  in  circumference  and 

6  ft.  high.    How  many  bushels  does  it  contain,  if  a  bushel  is  lieu,  ft.f 

y  Ex.  9.  What  fraction  of  a  pint  will  a  conical  wine-glass  hold,  if 
its  altitude  is  3  in.  and  the  diameter  of  the  top  ia  2  in.  ? 


NUMERICAL    EXERCISES    IN    SOLID    GEOMETRY     477 

Ex.  10.    Find  the  ratio  of  the  volumes  of  the  two  cones  inscribed 
in,  and  circumscribed  about,  a  regular  tetrahedron. 

Ex.  11.   If  an  equilateral  cone  contains  1  quart,  find  its  dimen- 
sions in  inches. 

Ex.  12.    In  a  cone  of  revolution  find  V  in  terms  of  R  and  L  ;  also 
find  V  in  terms  of  R  and  S. 

Ex.  13.    Find  the  volume  of  a  frustum  of  a  cone  of  revolution, 
whose  radii  are  14  and  7  ft.,  and  whose  altitude  is  2  yds. 

'  Ex.  14.  What  is  the  cost,  at  50  cts.  a  cu.  ft.,  of  a  piece  of  marble 
in  the  shape  of  a  frustum  of  a  cone  of  revolution,  whose  radii  are  6 
and  9  ft.,  and  whose  slant  height  is  5  ft.? 

Ex.  15.    In  a  frustum  of  a  cone  of  revolution,  the  volume  is  88 
cu.  ft.,  the  altitude  is  9  ft.,  and  R  =  2r.     Find  r. 


EXERCISES.     CROUP  8O 

SPHERICAL    VOLUMES 
Ex.  1.    Find  the  volume  of  a  sphere  whose  radius  is  1  ft.  9  in. 

Ex.  2.    If  the  earth  is  a  sphere  7,920  miles  in  diameter,  find  its 
***+***..  volume  in  cubic  miles. 

T' 

/      Ex.  3.    Find  the  diameter  of  a  sphere  whose  volume  is  1  cu.  ft. 

Ex.  4.    What  is  the  volume  of  a  sphere  whose  surface  is  616  sq.  in.  ? 


ML 

//^rty 


Ex.  5.    Find  the  radius  of  a  sphere  equivalent  to  the  sum  of  two 
spheres,  whose  radii  are  2  and  4  in. 

Ex.  6.    Find  the  radius  of  a  sphere  whose  volume  equals  the  area 
of  its  surface. 

Ex.  7.    Find  the  volume  of  a  sphere  circumscribed  about  a  cube 
whose  edge  is  6. 

•"  Ex.  8.    Find  the  volume    of   a  spherical   shell   whose   inner  and 
outer  diameters  are  14  and  21  in. 

Ex.  9.    Find  the  volume  of  a  spherical  shell  whose  inner  and  outer 
Surfaces  are  20  TT  and  12  TT. 


478  SOLID     GEOMETRY 

Find  the  volume  of 

Ex.  10.   A  spherical  wedge  whose  angle  is  24°,  the  radius  of  the 
sphere  being  10  in. 

-   Ex.  11.   A  spherical  sector  whose  base  is  a  zone  2  in.  high,  the 
radius  of  the  sphere  being  10  in. 


V 

"icallsc 


Ex.  12.    A  spherical  ^segment  of  two  bases  whose  radii  are  4  and 
7  and  altitude  5  in. 

'  Ex.  13.  A  wash-basin  in  the  shape  of  a  segment  of  a  sphere  is 
6  in.  deep  and  24  in.  in  diameter.  How  many  quarts  of  water  will  the 
basin  hold  ? 

/  Ex.  14.  A  plane  parallel  to  the  base  of  a  hemisphere  and  bisect- 
ing the  altitude  divides  its  volume  in  what  ratio  ? 

/  Ex.  15.  A  spherical  segment  4  in.  high  contains  200  cu.  in. ;  find 
the  radius  of  the  sphere. 

f  Ex.  16.  If  a  heavy  sphere  whose  diameter  is  4  in.  be  placed  in  a 
conical  wine-glass  full  of  water,  whose  diameter  is  5  in.  and  altitude 
6  in.,  find  how  much  water  will  run  over. 


EXERCISES.     GROUP  81 

EQUIVALENT    SOLIDS 

I  Ex.  1.  If  a  cubical  block  of  putty,  each  edge  of  which  is  8  inches, 
be  molded  into  a  cylinder  of  revolution  whose  radius  is  3  inches,  find 
the  altitude  of  the  cylinder. 

Ex.  2.  Find  the  radius  of  a  sphere  equivalent  to  a  cube  whose 
edge  is  10  in. 

-*, 

Ex.  3.  Find  the  radius  of  a  sphere  equivalent  to  a  cone  of  revolu- 
tion, whose  radius  is  3  in.  and  altitude  6  in. 

Ex.  4.  Find  the  edge  of  a  cube  equivalent  to  a  frustum  of  a  cone 
of  revolution,  whose  radii  are  4  and  9  ft.  and  altitude  2  yds. 

Ex.  5.  Find  the  altitude  of  a  rectangular  parallelepiped,  whose 
base  is  3  x5  in.  and  whose  volume  is  equivalent  to  a  sphere  of  radius 
7  in. 


NUMEKICAL    EXERCISES    IN    SOLID    GEOMETEY     479 

V  Ex.  6.  Find  the  base  of  a  square  rectangular  parallelepiped, 
whose  altitude  is  8  in.  and  whose  volume  equals  the  volume  of  a  cone 
of  revolution  with  a  radius  of  6  and  an  altitude  of  12  in. 

Ex.  7.  Find  the  radius  of  a  cone  of  revolution,  whose  altitude  is 
15  and  whose  volume  is  equal  to  that  of  a  cylinder  of  revolution  with 
radius  6  and  altitude  20. 


Ex.  8.  Find  the  altitude  of  a  cone  of  revolution,  whose  radius  is 
15  and  whose  volume  equals  the  volume  of  a  cone  of  revolution  with 
radius  9  and  altitude  24. 


X. 


Ex.  9.  On  a  sphere  whose  diameter  is  14  the  altitude  of  a  zone  of 
one  base  is  2.  Find  the  altitude  of  a  cylinder  of  revolution,  whose 
base  equals  the  base  of  the  zone  and  whose  lateral  surface  equals  the 
surface  of  the  zone. 


EXERCISES.     GROUP  82 

SIMILAR    SOLIDS 


Ex.  1.  If  on  two  similar  solids  L,  L'  and  I,  V  are  pairs  of  homol- 
ogous lines;  Af  A1  and  a,  a'  pairs  of  homologous  areas,  V,  V  and 
v,  vf  pairs  of  homologous  volumes, 

CL  :L'  =  l  :  l'  =  VA  :  i/A'=&v  :  tfv'. 

show  that      A  :  A'  =  LZ  :  Ln  =  a  :  a'=  vl  :  F'i. 


Ex.  2.    If  the  edge  of  a  cube  is  10  in.,  find  the  edge  of  a  cube 
having  5  times  the  surface. 

Ex.  3.    If  the  radius  of  a  sphere  is  10  in.,  find  the  radius  of  a 
sphere  having  5  times  the  surface. 

Ex.  4.    If  the  altitude  of  a  cone  of  revolution  is  10  in.,  find  the 
altitude  of  a  similar  cone  of  revolution  having  5  times  the  surface. 

Ex.  5.    In  the  last  three  exercises,  find  the  required  dimension  if 
the  volume  is  to  be  5  times  the  volume  of  the  original  solid. 

Ex.  6.    The  linear  dimensions  of  one  trunk  are  twice  as  great  as 
those  of  another  trunk.  How  much  greater  is  the  faunae  ?  Tbe  surface? 


480  SOLID     GEOMETRY 

Ex.  7.  How  far  from  the  vertex  is  the  cross -section  which  bisects 
the  volume  of  a  cone  of  revolution  ?  Which  bisects  the  lateral 
surface  ? 

Ex.  8.  If  the  altitude  of  a  pyramid  is  bisected  by  a  plane  parallel 
to  the  base,  how  does  the  area  of  the  cross-section  compare  with  the 
area  of  the  base  ?  How  does  the  volume  cut  off  compare  with  the 
volume  of  the  entire  pyramid  ? 

Ex.  9.  Planes  parallel  to  the  base  of  a  cone  divide  the  altitude 
into  three  equal  parts ;  compare  the  lateral  surfaces  cut  off.  Also  the 
volumes. 

Ex.  10.  A  sphere  10  in.  in  diameter  is  divided  into  three  equiva- 
lent parts  by  concentric  spherical  surfaces.  Find  the  diameters  of 
these  surfaces. 

Ex.  11.  If  the  strength  of  a  muscle  is  as  the  area  of  its  cross- 
section,  and  Goliath  of  Gath  was  three  times  as  large  in  each  linear 
dimension  as  Tom  Thumb,  how  much  greater  was  his  strength  ?  His 
weight  ?  How,  then,  does  the  activity  of  the  one  man  compare  with 
that  of  the  other  ? 

Ex.  12.  If  the  rate  at  which  heat  radiates,  from  a  body  is  in  pro- 
portion to  the  amount  of  surface,  and  the  planet  Jupiter  has  a  diame- 
ter 11  times  that  of  the  earth,  how  many  times  longer  will  Jupiter  be 
in  cooling  off  ? 

[Suo.  How  many  times  greater  is  the  volume,  and  therefore  .the 
original  amount  of  heat  in  Jupiter  ?  How  many  times  greater  is  its 
surface  ?  What  will  be  the  combined  effect  of  these  factors  ?] 


CROUP  83 

MISCELLANEOUS    NUMERICAL    EXERCISES    IN    SOLID 

GEOMETRY 
Find  S,  T  and  V  of 

Ex.  1.    A  right  triangular  prism  whose  altitude  is  1  ft.,  and  the 
sides  of  whose  base  are  26,  28,  30  in. 

Ex.  2.    A  cone  .of  revolution  the  radius  of  whose  base  is  1  ft.  2  in., 
and  whose  altitude  is  35  in. 

Ex.  3.    A  frustum  of  a  square  pyramid  the  areas  of  whose  bases 
are  1  sq,  ft.  and  36  sq,  in.,  and  whose  altitude  is  9  in. 


NUMEKICAL    EXEECISES    IN    SOLID    GEOMETEY     481 

Ex.  4.  A  pyramid  whose  slant  height  is  10  in.,  and  whose  base 
is  an  equilateral  triangle  whose  side  is  8  in. 

k     Ex.  5.    A  cube  whose  diagonal  is  1  yd. 

%  Ex.  6.  A  frustum  of  a  cone  of  revolution  whose  radii  are  6  and 
11  in.  and  slant  height  13  in. 

/^  Ex.  7.  A  rectangular  parallelepiped  whose  diagonal  is  2^29,  and 
whose  dimensions  are  in  the  ratio  2  :  3  :  4. 

*    Ex.  8.    Find  the  volume  of  a  sphere  inscribed  in  a  cube  whose 
'edge  is  6 ;  also  find  the  area  of  a  triangle  on  that  sphere  whose  angles 
are  80°,  90°,  150°. 

X  Ex.  9.  Find  the  volume  of  the  spherical  pyramid  whose  base  is 
the  above  triangle. 

X  Ex.  10.  Find  the  angle  of  a  lune  on  the  same  sphere,  equivalent 
to  that  triangle. 

£  Ex.  11.  On  a  cube  whose  edge  is  4,  planes  through  the  midpoints 
of  the  edges  cut  off  the  corners.  Find  the  volume  of  the  solid  re- 
maining. 

Ex.  12.  How  is  V  changed  if  H  of  a  cone  of  revolution  is  doubled 
and  R  remains  unchanged  ?  If  R  is  doubled  and  H  remains  unchanged? 
If  both  H  and  B  are  doubled  ? 

jL   Ex,  13.    In  an  equilateral  cone,  find  S  and  V  in  terms  of  E. 

Ex.  14.  A  piece  of  lead  20x8  x2  in.  will  make  how  many  spher- 
ical bullets,  each  f  in.  in  diameter  ? 

Ex.  15.  How  many  bricks  are  necessary  to  make  a  chimney  in 
the  shape  of  a  frustum  of  a  cone,  whose  altitude  is  90  ft.,  whose  outer 
diameters  are  3  and  8  ft.,  and  whose  inner  diameters  are  2  and  4  ft., 
counting  12  bricks  to  the  cubic  ft.? 

Ex.  16.  If  the  area  of  a  zone  is  300  and  its  altitude  6,  find  the 
radius  of  the  sphere. 

Ex.  17.  If  the  section  of  a  cylinder  of  revolution  through  its  axis 
is  a  square,  find  S,  T,  V  in  terms  of  E. 

Ex.  18.    If  every  edge  of  a  square  pyramid  is  fy  find  b  in  terms  of  T. 
DD 


452  SOLID    GEOMETRY 

^  Ex.19.  A  regular  square  pyramid  has  a  for  its  altitude  and  also  for 
each  side  of  the  base.  Find  the  area  of  a  section  made  by  a  plane  parallel 
to  the  base  and  bisecting  the  altitude.  Find  also  the  volumes  of  the 
two  parts  into  which  the  pyramid  is  divided. 

/  Ex.  20.  If  the  earth  is  a  sphere  of  8,000  miles  diameter  and  its 
atmosphere  extends  50  miles  from  the  earth,  find  the  volume  of  the 
atmosphere. 

Ex.  21.  On  a  sphere,  find  the  ratio  of  the  area  of  an  equilateral 
spherical  triangle,  each  of  whose  angles  is  95°,  to  the  area  of  a  lune 
whose  angle  is  80°. 

Ex.  22.  A  square  right  prism  has  an  altitude  6a  and  an  edge  of 
the  base  2a.  Find  the  volume  of  the  largest  cylinder,  sphere,  pyramid 
and  cone  which  can  be  cut  from  it. 

I/    Ex.  23.    Obtain  a  formula  for  the  area  of  that  part  of  a  sphere 
'illuminated   by  a    point    of   light  at  a  distance    a  from  tne    sphere 
whose  radius  is  E. 

*/     Ex.  24.    On  a  sphere  whose  radius  is  6  in.,  find  an  angle  of  an 
/  equilateral  triangle  whose  area  is  12  sq.  in. 

Ex.  25.  Find  the  volume  of  a  prismatoid,  whose  altitude  is  24  and 
whose  bases  are  equilateral  triangles,  each  side  10,  so  placed  that  the 
mid-section  of  the  prismatoid  is  a  regular  hexagon. 

/  Ex.  26.  On  a  sphere  whose  radius  is  16,  the  bases  of  a  zone  are 
equal  and  are  together  equal  to  the  area  of  the  zone.  Find  the  alti- 
tude of  the  zone. 

/•"  Ex.  27.  Find  the  volume  of  a  square  pyramid,  the  edge  of  whose 
base  is  10  and  each  of  whose  lateral  edges  is  inclined  60°  to  the  base. 

^  Ex.  28.  An  irregular  piece  of  ore,  if  placed  in  a  cylinder  partly 
filled  with  water,  causes  the  water  to  rise  6  in.  If  the  radius  of  the 
cylinder  is  8  in.,  what  is  the  volume  of  the  ore  ? 

Ex.  29.  Find  the  volume  of  a  truncated  right  triangular  prism,  if 
the  edges  of  the  base  are  8,  9,  11,  and  the  lateral  edges  are  12,  13,  14. 


-„ 


Ex.  30.  In  a  sphere  whose  radius  is  5,  a  section  is  taken  at  tho 
distance  3  from  the  center.  On  this  section  as  a  base  a  cone  is  formed 
whose  lateral  elements  are  tangent  to  the  sphere,  Find  tho  lateral 
surface  and  volume  of  the  cone. 


NUMEKICAL    EXERCISES    IN    SOLID    GEOMETRY     483 

Ex.  31 .    The  volume  of  a  sphere  is  l,437i  cu.  in.    Find  the  surface. 

Ex.  32.  A  square  whose  side  is  6  is  revolved  about  a  diagonal  as 
an  axis;  find  the  surface  and  volume  generated. 

Ex.  33.  Find  the  edge  of  a  cubical  cistern  that  will  hold  10  tons 
of  water,  if  1  cu.  ft.  of  water  weighs  62.28  Ibs. 

Ex.  34.  A  water  trough  has  equilateral  triangles,  each  side  3  ft., 
for  ends,  and  is  18  ft.  long.  How  many  buckets  of  water  will  it  hold, 
if  a  bucket  is  a  cylinder  1  ft.  in  diameter  and  li  ft.  high  ? 

Ex.  35.  The  lateral  area  of  a  cylinder  of  revolution  is  440  sq.  in., 
and  the  volume  is  1,540  cu.  in.  Find  the  radius  and  altitude. 

Ex.  36.  The  angles  of  a  spherical  quadrilateral  are  80°,  100°, 
120°,  120°.  Find  the  angle  of  an  equivalent  equilateral  triangle. 

if      Ex.  37.    A  cone  and  a  cylinder  have  equal  lateral  surfaces,  and 
their  axis  sections  are  equilateral.    Find  the  ratio  of  their  volumes. 

Ex.  38.  A  water-pipe  £  in.  in  diameter  rises  13  f.t.  from  the  ground. 
How  many  quarts  of  water  must  be  drawn  from  it  before  the  water 
from  under  the  ground  comes  out  ?  If  a  quart  runs  out  in  5  seconds, 
how  long  must  the  water  run  ? 

Ex.  39.  A  cube  immersed  in  a  cylinder  partly  filled  with  water 
causes  the  water  to  rise  4  in.  If  the  radius  of  the  cylinder  is  6  in., 
what  is  an  edge  of  the  cube  ? 

Ex.  40.  An  auger  hole  whose  diameter  is  3  in.  is  bored  through 
the  center  of  a  sphere  whose  diameter  is  8  inches.  Find  the  volume 
remaining. 

Ex.  41.  Show  that  the  volumes  of  a  cone,  hemisphere,  and  cylin- 
der of  the  same  base  and  altitude  are  as  1  '.  2  :  3. 

Ex.  42.  The  volumes  of  two  similar  cylinders  of  revolution  are 
as  8  :  125 ;  find  the  ratio  of  their  radii.  If  the  radius  of  the  smaller  is 
10  in.,  what  is  the  radius  of  the  larger  ? 

Ex.  43.  An  iron  shell  is  2  in.  thick  and  the  diameter  of  its  outer 
surface  is  28  in.  Find  its  volume. 

Ex.  44.  The  legs  of  an  isosceles  spherical  triangle  each  make  an 
angle  of  75°  with  the  base.  The  legs  produced  form  a  lune  whose 
area  is  four  times  the  area  of  the  triangle.  Find  the  angle  of  the  lune. 


484  SOLID     GEGMETEY 

CROUP  84 

EXERCISES    INVOLVING    THE    METRIC    SYSTEM 
Find  S,  T,  V  of 

Ex.  1.    A  right  prism  the  edges  of  whose  base  are  6m.,  70  dm., 
900  cm.,  and  whose  altitude  is  90  dm. 

Ex.  2.    A  regular  square  pyramid  an  edge  of  whose  base  is  30  dm., 
and  whose  altitude  is  1.7  mP 

Ex.  3.    A  sphere  whose  radius  is  0.02  m. 

Ex.  4.    A  frustum  of  a  cone  of  revolution  whose  radii  are  10  dm. 
and  6  dm.,  and  whose  slant  height  is  50  cm. 

Ex.  5.    A  cube  whose  diagonal  is  12  cm. 

Ex.  6.    A  cylinder  of  revolution  whose  radius  equals  2  dm.,  and 
whose  altitude  equals  the  diameter  of  the  base. 

Ex.  7.    Find  the  area  of  a  spherical  triangle  on  a  sphere  whose 
radius  is  0.02  m.,  if  its  angles  are  110°,  120°,  130°. 

Ex.  8.    Find  the  number  of  square  meters    in   the  surface  of    a 
sphere,  a  great  circle  of  which  is  50  dm.  long. 

Ex.  9.    How  many    liters  will  a   cylindrical   vessel    hold   that  is 
10  dm.  in  diameter  and  0.25  m.  high  ?    How  many  liquid  quarts  ? 

Ex.  10.    A  liter  measure  is  a  cylinder  whose  diameter  is  half  the 
altitude.    Find  its  dimensions  in  centimeters. 

Ex.  11.    Find  the  surface  of  a  sphere  whose  volume  is  1  cu.  m,, 


APPENDIX 

J.    MODERN   GEOMETRIC   CONCEPTS 

840.  Modern   Geometry.     In   recent   times  many  new 
geometric  ideas  have  been    invented,   and  some  of  them 
developed    into    important    new    branches    of    geometry. 
Thus,  the   idea    of   symmetry    (see  Art.   484,   etc.)    is   a 
modern  geometric  concept.     A  few  other  of  these  modern 
concepts    and    methods    will    be    briefly    mentioned,    but 
their    thorough    consideration    lies  beyond    the   scope    of 
this  book. 

841.  Projective   Geometry.      The    idea   of   projections 
(see    Art.    345)   has    been    developed    in    comparatively 
recent  times  into  an  important  branch  of  mathematics  with 
many  practical  applications,  as  in  engineering,  architec- 
ture, construction  of  maps,  etc. 

842.  Principle  of  Continuity.      By   this   principle  two 
or  more  theorems  are  made  special  cases  of  a  single   more 
general  theorem.    An  important  aid  in  obtaining  continuity 
among  geometric  principles  is  the  application  of  the  con- 
cept of  negative  quantity  to  geometric  magnitudes. 

Thus,  a  negative  line  is  a  line  opposite  in  direction  to 
a  given  line  taken  as  positive. 


For  example,  if  OA  is  +,  OB  is  — , 

(485) 


486 


GEOMETKY.      APPENDIX 


Similarly,  a  negative  angle  is 
an  angle  formed  by  rotating  a  line 
in  a  plane  in  a  direction  opposite 
from  a  direction  of  rotation  taken 
as  positive.  Thus,  if  the  line  OA 
rotating  from  the  position  OA 
forms  the  positive  angle  AOB,  the 

same  line  rotating  in  the  opposite  direction  forms  the 
negative  angle  AOB'.  Similarly,  positive  and  negative 
arcs  are  formed. 

In  like  manner,  if  P  and  Pf  are 
on  opposite  sides  of  the  line  AB  and 
the  area  PAB  is  taken  as  positive, 
the  area  P'AB  will  be  negative. 

As  an  illustration  of  the  law  of 
continuity,  we  may  take  the  theorem 
that  the  sum  of  the  triangles  formed 
by  drawing  lines  from  a  point  to  the 
vertices  of  a  polygon  equals  the  area 
of  the  polygon. 

Applying  this  to  the  quadrilateral  ABCD,  if  the  point 
P  falls  within  the  quadrilateral,  APA£+AP£C  +  APOD 
+  APAD=ABCD  (Ax.  6). 

Also,  if  the  point  falls  without  the  quadrilateral  at  P', 
A  P'AB  +  A  P'BC  +  A  P'CD  +  A  P'AD  =  ABCD,  since 
AP'AD  is  a  negative  area,  and  hence  is  to  be  subtracted 
from  the  sum  of  the  other  three  triangles. 

843.  The  Principle  of  Reciprocity,  or  Duality,  is  a 
principle  of  relation  between  two  theorems  by  which  each 
theorem  is  convertible  into  the  other  by  causing  the  words 
for  the  same  two  geometric  objects  in  each  theorem  to  ex- 
change places. 


MODERN     GEOMETRIC     CONCEPTS 


487 


Thus,  of  theorems  VI  and  VII,  Book  I,  either  may  be 
converted  into  the  other  by  replacing  the  word  "sides"  by 
"angles,"  and  "angles"  by  "sides."  Hence  these  are 
termed  reciprocal  theorems. 

The  following  are  other  instances  of  reciprocal  geometric 
properties : 


1.  Two     points     determine     a 
straight  line. 

2.  Three  points  not  in  the  same 
straight  line  determine  a  plane. 

3.  A  straight  line  and  a  point 
determine  a  plane. 


1.  Two  lines  determine  a  point. 

2.  Tliree  planes  not  through  the 
same    straight    line     determine    a 
point. 

3.  A  straight  line  and  a  plane 
determine  a  point. 


The  reciprocal  of  a  theorem  is  not  necessarily  true. 

Thus,  two  parallel  straight  lines  determine  a*plane,  but 
two  parallel  planes  do  not  determine  a  line. 

However,  by  the  use  of  the  principle  of  reciprocity, 
geometrical  properties,  not  otherwise  obvious,  are  fre- 
quently suggested. 

844.  Principle  of  Homology.  Just  as  the  law  of  reci- 
procity indicates  relations  between  one  set  of  geometric 
concepts  (as  lines)  and  another  set  of  geometric  concepts 
(as  points),  so  the  law  of  homology  indicates  relations 
between  a  set  of  geometric  concepts  and  a  set  of  concepts 
outside  of  geometry:  as  a  set  of  algebraic  concepts,  for 
instance. 

Thus,  if  a  and  6  are  numbers,  by  algebra  (a  +  5)  (a — 5) 
=  a2— 62. 

Also,  if  a  and  b  are  segments  of  a  line,  the  rectangle 
(a-j-ft)X(a — &)  is  equivalent  to  the  difference  between  the 
squares  a2  and  62. 

By  means  of  this  principle,  truths  which  would  be  over- 
looked or  difficult  to  prove  ill  one  department  of  thought 


488  GEOMETRY.      APPENDIX 

are  made  obvious  by  observing  the  corresponding  truth  in 
another  department  of  thought. 

Thus,  if  a  and  I  are  line  segments,  the  theorem  (a+  6)2 
+  (a  —  &)2=2(a2-f-62)  is  not  immediately  obvious  in  geo- 
metry, but  becomes  so  by  observing  the  like  relation 
between  the  algebraic  numbers  a  and  &. 

i 

845.  Non-Euclidean  Geometry.     Hyperspace.     By  vary- 
ing the  properties  of  space,  as  these  are  ordinarily  stated, 
different  kinds  of  space  may  be  conceived  of,  each  having 
its  own  geometric   laws   and   properties.    Thus,  space,  as 
we  ordinarily  conceive  it,  has  three  dimensions,  but  it  is 
possible    to    conceive    of  space    as   having  four  or   more 
dimensions.     To  mention  a  single  property  of  four  dimen- 
sional  space,  in  such  a  space  it  would    be   possible,  by 
simple    pressure,  to  turn   a  sphere,  as    an  orange,  inside 
out  without  breaking  its  surface. 

As  an  aid  toward  conceiving  how  this  is  possible,  consider  a  plane 
in  which  one  circle  lies  inside  another.  No  matter  how  these  circles 
are  moved  about  in  the  plane,  it  is  impossible  to  shift  the  inner  circle 
so  as  to  place  it  outside  the  other  without  breaking  the  circumference 
of  the  outer  circle.  But,  if  we  are  allowed  to  use  the  third  dimension 
of  space,  it  is  a  simple  matter  to  lift  the  inner  circle  up  out  of  the 
plane  and  set  it  down  outside  the  larger  circle. 

Similarly  if,  in  space  of  three  dimensions,  we  have  one  spherical 
shell  inside  a  larger  shell,  it  is  impossible  to  place  the  smaller  shell 
outside  the  larger  without  breaking  the  larger.  But  if  the  use  of  a 
fourth  dimension  be  allowed,— that  is,  the  use  of  another  dimension 
of  freedom  of  motion,— it  is  possible  to  place  the  inner  shell  outside 
the  larger  without  breaking  the  latter. 

846.  Curved  Spaces.     By  varying  the  geometric  axioms 
of  space   (see  Art.  '47),   different   kinds  of  space  may  be 
conceived  of.     Thus,  we  may  conceive  of  space  such  that 
through  a  given  point  one  line  may  be  drawn  parallel  to  a 
given  line  (that  is  ordinary,  or  Euclidean  space);  or  such 


MODEKN     GEOMETRIC     CONCEPTS  489 

that  through  a  given  point  no  line  can  be  drawn  parallel 
to  a  given  line  (spherical  space) ;  or  such  that  through  a 
given  point  more  than  one  line  can  be  drawn  parallel  to  a 
given  line  (pseudo- spherical  space). 

These  different  kinds  of  space  differ  in  many  of  their 
properties.  For  example,  in  the  first  of  them  the  sum  of 
the  angles  of  a  triangle  equals  two  right  angles;  in  the 
second,  it  is  greater;  in  the  third,  it  is  less. 

These  different  kinds  of  space,  however,  have  many 
properties  in  common.  Thus,  in  all  of  them  every  point  in 
the  perpendicular  bisector  of  a  line  is  equidistant  from  the 
extremities  of  the  line. 


EXERCISES.     GROUP  85 

Ex.  1.  Show  by  the  use  of  zero  and  negative  arcs  that  the  princi- 
ples of  Arts.  257,  263,  258,  264,  265,  are  particular  cases  of  the  general 
theorem  that  the  angle  included  between  two  lines  which  cut  or  touch 
a  circle  is  measured  by  one-half  the  sum  of  the  intercepted  arcs. 

Ex.  2.  Show  that  the  principles  of  Arts.  354  and  358  are  particular 
cases  of  the  theorem  that,  if  two  lines  are  drawn  from  or  through  a 
point  to  meet  a  circumference,  the  product  of  the  segments  of  one  line 
equals  the  product  of  the  segments  of  the  other  line. 

Ex.  3.  Show  by  the  use  of  negative  angles 
that  theorem  XXXVIII,  Book  I,  is  true  for  a 
quadrilateral  of  the  form  ABCD.  [BCD  is  a 
negative  angle  ;  the  angle  at  the  vertex  D  is 
the  reflex  angle 


Ex.  4.  What  is  the  reciprocal  of  the  state- 
ment that  two  intersecting  straight  lines  deter- 
mine a  plane  ? 

Ex.  5.  What  is  the  reciprocal  of  the  statement  that  three  planes 
perpendicular  to  each  other  determine  three  straight  lines  perpen' 
dicular  to  each  other  ? 


H.    HISTORY   OF   GEOMETRY 

847.  Origin  of  Geometry  as  a  Science.    The  beginnings 
of  geometry  as  a  science  are  found  in  Egypt,  dating  back 
at  least  three  thousand  years   before  Christ.     Herodotus 
says  that  geometry,  as  known  in  Egypt,  grew  out  of  the 
need  of  remeasuring  pieces  of  land  parts  of  which  had  been 
washed  away  by  the  Nile  floods,  in  order  to  make  an  equi- 
table readjustment  of  the  taxes  on  the  same. 

The  substance  of  the  Egyptian  geometry  is  found  in  an 
old  papyrus  roll,  now  in  the  British  museum.  This  roll 
is,  in  effect,  a  mathematical  treatise  written  by  a  scribe 
named  Ahmes  at  least  1700  B.C.,  and  is,  the  writer  states, 
a  copy  of  a  more  ancient  work,  dating,  say,  3000  B.  C. 

848.  Epochs  in  the  Development  of  Geometry.     From 
Egypt  a  knowledge  of  geometry  was  transferred  to  Greece, 
whence  it  spread  to  other  countries.     Hence  we  have  the 
following  principal  epochs  in  the  development  of  geometry; 

1.  Egyptian  :  3000  B.  C.— 1500  B.  C. 

2.  Greek  :  600  B.  C.— 100  B.  C. 

3.  Hindoo  :  500  A.  D.— 1100  A.  D. 

4.  Arab  :  800  A.  D.— 1200  A.  D. 

5.  European  :  1200  A.  D. 

In  the  year  1120  A.  D.,  Athelard,  an  English  monk, 
visited  Cordova,  in  Spam,  in  the  disguise  of  a  Mohamme- 
dan student,  and  procured  a  copy  of  Euclid  m  the  Arabic 
language.  This  book  he  brought  back  to  central  Europe, 
where  it  was  translated  into  Latin  and  became  the  basis  of 
all  geometric  study  m  Europe  till  the  year  1633,  when, 

(490) 


HISTORY    OF     GEOMETRY  491 

owing  to  the  capture  of  Constantinople  by  the  Turks, 
copies  of  the  works  of  the  Greek  mathematicians  in  the 
original  Greek  were  scattered  through  Europe. 


HISTORY   OF   GEOMETRICAL    METHODS 

-•*.        f- 

849.  Rhetorical  Methods.  By  rhetorical  methods  in 
the  presentation  of  geometric  truths,  is  meant  the  use  of 
definitions,  axioms,  theorems,  geometric  figures,  the  rep- 
resentation of  geometric  magnitudes  by  the  use  of  letters, 
the  arrangement  of  material  in  Books,  etc.  The  Egyptians 
had  none  of  these,  their  geometric  knowledge  being  re- 
corded only  in  the  shape  of  the  solutions  of  certain  numeri- 
cal examples,  from  which  the  rules  used  must  be  inferred. 

Thales  (Greece  600  B.C.)  first  made  an  enunciation  of 
an  abstract  property  of  a  geometric  figure.  He  had  a  rude 
Idea  of  the  geometric  theorem. 

Pythagoras  (Italy  525,  B.C.)  introduced  formal  defini- 
tions into  geometry,  though  some  of  those  used  by  him 
were  not  very  accurate.  F,or  instance,  his  definition  of  a 
point  is  "unity  having  position."  Pythagoras  also 
arranged  the  leading  propositions  known  to  him  in 
something  like  logical  order. 

Hippocrates  (Athens,  420  B.  C.)  was  the  first  systemati- 
cally to  denote  a  point  by  a  capital  letter,  and  a  segment 
of  a  line  by  two  capital  letters,  as  the  line  AB,  as  is  done 
at  present.  He  also  wrote  the  first  text  -book  on  geometry. 

Plato  (Athens,  380  B.  C.)  made  definitions,  axioms 
and  postulates  the  beginning  and  basis  of  geometry. 

To  Euclid  (Alexandria,  280  B.  C.)  is  due  the  division 
of  geometry  into  Books,  the  formal  enunciation  of  theo- 
rems, the  particular  enunciation,  the  formal  construction, 


492  GEOMETKY.      APPENDIX 

proof,  and  conclusion,  in  presenting  a  proposition.  He 
also  introduced  the  use  of  the  corollary  and  scholium. 

Using  these  methods  of  presenting  geometric  truths, 
Euclid  wrote  a  text -book  of  geometry  in  thirteen  books, 
which  was  the  standard  text -book  on  this  subject  for 
nearly  two  thousand  years. 

The  use  of  the  symbols  A,  ZZ7  ,  ||  ,  etc.,  in  geometric 
proofs  originated  in  the  United  States  in  recent  years. 

850.  Logical  Methods.  The  Egyptians  used  no  formal 
methods  of  proof.  They  probably  obtained  their  few  crude 
geometric  processes  as  the  result  of  experiment. 

The  Hindoos  also  used  no  formal  proof.  One  of  their 
writers  on  geometry  merely  states  a  theorem,  draws  a 
figure,  and  says  "Behold  !  " 

The  use  of  logical  methods  of  geometric  proof  is  due  to 
the  Greeks.  The  early  Greek  geometricians  used  experi- 
mental methods  at  times,  in  order  to  obtain  geometric 
truths.  For  instance,  they  determined  that  the  angles  at 
the  base  of  an  isosceles  triangle  are  equal,  by  folding  half 
of  the  triangle  over  on  the  altitude  as  an  axis  and  observ- 
ing that  the  angles  mentioned  coincided  as  a  fact,  but 
without  showing  that  they  must  coincide. 

Pythagoras  (525  B.  C.)  was  the  first  to  establish  geo- 
metric truths  by  systematic  deduction,  but  his  methods 
were  sometimes  faulty.  For  instance,  he  believed  that  the 
converse  of  a  proposition  is  necessarily  true. 

Hippocrates  (420  B.C.)  used  correct  and  rigorous  de- 
duction in  geometric  proofs.  He  also  introduced  specific 
varieties  of  such  deduction,  such  as  the  method  of  reduc- 
ing one  proposition  to  another  (Art.  296),  and  the  reductio 
ad  absurdum. 


HISTORY    OF     GEOMETRY  493 

The  methods  of  deduction  used  by  the  Greeks,  however,  were  de- 
fective in  their  lack  of  generality.     For  instance,  it  was  often  thought 
necessary  to  have  a  separate  proof  of  a  theorem  for  each  different 
kind  of  figure  to  which  the  theorem  applied. 
Thus,  the  theorem  that  the  sum  of  the  an- 
gles of  a  triangle  equals  two  right  angles 
was  proved, 

(1)  for  the  equilateral   triangle  by  use 
of  the  regular  hexagon ; 


(2)  for  the  right  triangle  by  the  use  of 


a  rectangle ; 

(3)  for  a  scalene  triangle  by  dividing 
the  scalene  triangle  into  two  right  triangles. 

The  Greeks  appeared  to  fear  that  a 
general  proof  might  be  vitiated  if  it  were 
applied  to  a  figure  in  any  way  special  or 

peculiar.    In  other  words,  they  had  no  conception  of  the  principle  of 
continuity  (Art.  842). 

Plato  (380  B.  C.)  introduced  the  method  of  proof  by 
analysis,  that  is,  by  taking  a  proposition  as  true  and  work- 
ing from  it  back  to  known  truths  (see  Art.  196). 

To  Eudoxus  (360  B.  C.)  is  virtually  due  proof  by  the 
method  of  limits;  though  his  method,  known  as  the 
method  of  exhaustions,  is  crude  and  cumbersome. 

Apollonius  (Alexandria,  225  B.  C.)  used  projections, 
transversals,  etc.,  which,  in  modern  times,  have  developed 
into  the  subject  of  projective  geometry. 

851.  Mechanical  Methods.  The  Greeks,  in  demonstra- 
ting a  geometrical  theorem,  usually  drew  the  figure  em- 
ployed in  a  bed  of  sand.  This  method  had  certain  advan- 
tages, but  was  not  adapted  to  the  use  of  a  large  audience. 

At  the  time  when  geometry  was  being  developed  in  Greece,  the 
interest  in  the  subject  was  very  general.  There  was  scarcely  a  town 
but  had  its  lectures  on  the  subject.  The  news  of  the  discovery  of  a 


494  GEOMETKY.      APPENDIX 

new  theorem  spread  from  town  to  town,  and  the  theorem  was  redemon- 
strated  in  the  sand  of  each  market  place. 

The  Greek  treatises,  however,  were  written  on  vellum 
or  papyrus  by  the  use  of  the  reed,  or  calamus,  and  ink. 

In  Roman  times,  and  in  the  middle  ages,  geometrical 
figures  were  drawn  in  wax  smeared  on  wooden  boards, 
called  tablets.  They  were  drawn  by  the  use  of  the  stylus, 
a  metal  stick,  pointed  at  one  end  for  making  marks,  and 
broad  at  the  other  for  erasing  marks.  These  wax  tablets 
were  still  in  use  in  Shakespeare's  time  (see  Hamlet  Act  I, 
Sc.  5,  1.  107).  The  blackboard  and  crayon  are  modern 
inventions,  their  use  having  developed  within  the  last  one 
hundred  years. 

The  Greeks  invented  many  kinds  of  drawing  instruments 
f6r  tracing  various  curves.  It  was  due  to  the  influence  of 
Plato  (380  B.  C.)  that,  in  constructing  geometric  figures, 
the  use  of  only  the  ruler  and  compasses  is  permitted. 


HISTORY  OF  GEOMETRIC   TRUTHS.     PLANE   GEOMETRY 

852.  Rectilinear  Figures.  The  Egyp- 
tians measured  the  area  of  any  four- 
sided  field  by  multiplying  half  the  sum 
of  one  pair  of  opposite  sides  by  half  the 
sum  of  the  other  pair;  which  was  equivalent  to  using  the 

a~rc^.b-\-d 
formula,  area  =    0    X    0    • 

_  — 

This,  of  course,  gives  a  correct  result  for  the  rectangle  and  square, 
but  gives  too  great  a  result  for  other  quadrilaterals,  as  the  trapezoid, 
etc.  Hence  Joseph,  of  the  Book  of  Genesis,  in  buying  the  fields  of 
the  Egyptians  for  Pharoah  in  time  of  famine  by  the  use  of  this 
formula,  in  many  cases  paid  for  a  larger  field  than  he  obtained. 

The  Egyptians  had  a  special  fondness  for  geometrical 
constructions,  probably  growing  out  of  their  work  as  temple 


HISTORY     OF    GEOMETRY  495 

builders.  A  class  of  workers  existed  among  them  called 
"rope -stretchers,"  whose  business  was  the  marking  out  of 
the  foundations  of  buildings.  These  men  knew  how  to 
bisect  an  angle  and  also  to  construct  a  right  angle.  The 
latter  was  probably  done  by  a  method  essentially  the  same 
as  forming  a  right  triangle  whose  sides  are  three,  four  and 
five  units  of  length.  Ahmes,  in  his  treatise,  has  various 
constructions  of  the  isosceles  trapezoid  from  different  data. 

Thales  (600  B.  C.)  enunciated  the  following  theorems: 

If  two  straight  lines  intersect,  the  opposite  or  vertical 
angles  are  equal; 

The  angles  at  the  base  of  an  isosceles  triangle  are  equal; 

Two  triangles  are  equal  if  two  sides  and  the  included 
angle  of  one  are  equal  to  two  sides  and  the  included  angle 
of  the  other; 

The  sum  of  the  angles  of  a  triangle  equals  two  right 
angles ; 

Two  mutually  equiangular  triangles  are  similar. 

Thales  used  the  last  of  these  theorems  to  measure  the 
height  of  the  great  pyramid  by  measuring  the  length  of 
the  shadow  cast  by  the  pyramid  and  also  measuring  the 
length  of  the  shadow  of  a  post  of  known  height  at  the  same 
time  and  making  a  proportion  between  these  quantities. 

Pythagoras  (525  B.  C.)  and  his  followers  discovered 
correct  formulas  for  the  areas  of  the  principal  rectilinear 
figures,  and  also  discovered  the  theorems  that  the  areas  of 
similar  polygons  are  as  the  squares  of  their  homologous 
sides,  and  that  the  square  on  the  hypotenuse  of  a  right 
triangle  equals  the  sum  of  the  squares  on  the  other  two 
sides.  The  latter  is  called  the  Pythagorean  theorem. 
They  also  discovered  how  to  construct  a  square  equivalent 
to  a  given  parallelogram,  and  to  divide  a  given  line  in 
mean  and  extreme  ratio, 


496  GEOMETRY.      APPENDIX 

To  Eudoxus  (380  B.  C.)  we  owe  the  general  theory  of 
proportion  in  geometry,  and  the  treatment  of  incommen- 
surable quantities  by  the  method  of  Exhaustions.  By  the 
use  of  these  he  obtained  such  theorems  as  that  the  areas 
of  two  circles  are  to  each  other  as  the  squares  of  their 
radii,  or  of  their  diameters. 

In  the  writings  of  Hero  (Alexandria,  125  B.  C.)  we  first 
find  the  formula  for  the  area  of  a  triangle  in  terms  of  its 

sides,  K=V  s(s  —  a)  (s  —  b)  (s  —  c)  -    Hero  also  was  the  first 
to  place  land-surveying  on  a  scientific  basis,, 

It  is  a  curious  fact  that  Hero  at  the  same  time  gives  an  incorrect 
formula  for  the  area  of  a  triangle,  viz.,  lT=ia(&-f-c),  this  formula 
being  apparently  derived  from  Egyptian  sources. 

Xenodorus  (150  B.  C.)  investigated  isoperemetrical 
figures. 

The  Romans,  though  they  excelled  in  engineering,  ap- 
parently did  not  appreciate  the  value  of  the  Greek  geom- 
etry. Even  after  they  became  acquainted  with  it,  they 
continued  to  use  antiquated  and  inaccurate  formulas  for 
areas,  some  of  them  of  obscure  origin.  Thus,  they  used 
the  Egyptian  formula  for  the  area  of  a  quadrilateral, 

K=^-  X—  —  .    They  determined  the  area  of  an  equilat- 

JJ          — 

eral  triangle  whose  side  is  a,  by  different  formulas,  all 
incorrect,  as  K=-r  ,  £T=i(a2+a),  and  j£=Ja2. 


853.  The  Circle.  Thales  enunciated  the  theorem  that 
every  diameter  bisects  a  circle,  and  proved  the  theorem 
that  an  angle  inscribed  in  a  semicircle  is  a  right  angle. 

To  Hippocrates  (420  B.  C.)  is  due  the  discovery  of 
nearly  all  the  other  principal  properties  of  the  circle  given 
in  this  book. 


HISTORY   OF   GEOMETRY  497 

The  Egyptians  regarded  the  area  of  the  circle  as  equiva- 
lent to  ff  of  the  diameter  squared,  which  would   make 


The  Jews  and  Babylonians  treated  n  as  equal  to  3. 

Archimedes,  by  the  use  of  inscribed  and  circumscribed 
regular  polygons,  showed  that  the  true  value  of  n  lies 
between  3|  and  3yf;  that  is,  between  3.14285  and  3.1408. 

The  Hindoo  writers  assign  various  values  to  7t,  as  3,  3i, 
1/10,  and  Aryabhatta  (530  A.  D.)  gives  the  correct  ap- 
proximation, 3.1416.  The  Hindoos  used  the  formula 

1/2  _  1/4—  AB2  (^ee  Art.  468)  in  computing  the  numeri- 
cal value  of  71. 

Within  recent  times,  the  value  of  n  has  been  computed 
to  707  decimal  places. 

The  use  of  the  symbol  n  for  the  ratio  of  the  circum- 
ference of  a  circle  to  the  diameter  was  established  in 
mathematics  by  Euler  (Germany,  1750). 


HISTORY   OF   GEOMETRIC  TRUTHS.      SOLID    GEOMETRY 

854.  Polyhedrons.  The  Egyptians  computed  the  vol- 
umes of  solid  figures  from  the  linear  dimensions  of  such 
figures.  Thus,  Ahmes  computes  the  contents  of  an  Egyp- 
tian barn  by  methods  which  are  equivalent  to  the  use 

3c 
of  the  formula  V=aXbX— .    As  the  shape  of  these  barns 

is  not  known,  it  is  not  possible  to  say  whether  this  formula 
is  correct  or  not. 

Pythagoras  discovered,  or  knew,  all  the  regular  poly- 
hedrons except  the  dodecahedron.  These  polyhedrons  were 
supposed  to  have  various  magical  or  mystical  properties. 
Hence  the  study  of  them  was  made  very  prominent. 

FF 


498  GEOMETEY.      APPENDIX 

Hippasus  (470  B.C.)  discovered  the  dodecahedron,  but 
lie  was  drowned  by  the  other  Pythagoreans  for  boasting 
of  the  discovery. 

Eudoxus  (380  B.  C.)  showed  that  the  volume  of  a  pyra° 
mid  is  equivalent  to  one -third  the  product  of  its  base  by 
its  altitude. 

E.  F.  August  (Germany,  1849)  introduced  the  prisma- 
toid  formula  into  geometry  and  showed  its  importance. 

855.  The  Three  Round  Bodies.     Eudoxus  showed  that 
the  volume  of  a  cone  is  equivalent  to  one -third  the  area  of 
its  base  by  its  altitude. 

Archimedes  discovered  the  formulas  for  the  surface  and 
volume  of  the  sphere. 

Menelaus  (100  A.  D.)  treated  of  the  properties  of 
spherical  triangles. 

Gerard  (Holland,  1620)  invented  polar  triangles  and 
found  the  formulas  for  the  area  of  a  spherical  triangle  and 
of  a  spherical  polygon. 

856.  Non-Euclidean  Geometry.     The  idea  that  a  space 
might  exist  having  different  properties  from  those  which 
we  regard  as  belonging  to  the  space  in  which  we  live,  has 
occurred    to    different    thinkers    at    different    times,   but 
Lobatchewsky  (Russia,  1793-1856)   was  the  first  to  make 
systematic  use  of  this  principle.     He  found  that  if,  instead 
of  taking  Geom.  Ax.  2  as  true,  we  suppose  that  through  a 
given  point  in  a  plane  several  straight  lines  may  be  drawn 
parallel  to  a  given  line,  the  result  is  not  a  series  of  absur- 
dities or  a  general  reductio  ad  absurdum;   but,  on  the  con- 
trary, a  consistent  series  of  theorems  is  obtained  giving 
the  properties  of  a  space. 


III.    REVIEW  EXERCISES 

EXERCISES.     GROUP  86 

REVIEW    EXERCISES    IN    PLANE    GEOMETRY 

Ex.  1.  If  the  bisectors  of  two  adjacent  angles  are  perpendiculai 
to  each  other,  the  angles  are  supplementary. 

Ex.  2.  If  a  diagonal  of  a  quadrilateral  bisects  two  of  its  angles, 
the  diagonal  bisects  the  quadrilateral. 

Ex.  3.  Through  a  given  point  draw  a  secant  at  a  given  distance 
from  the  center  of  a  given  circle. 

Ex.  4.  The  bisector  of  one  angle  of  a  triangle  and  of  an  exterior 
angle  at  another  vertex  form  an  angle  which  is  equal  to  one -half  the 
third  angle  of  the  triangle. 

Ex.  5.  The  side  of  a  square  is  18  in.  Find  the  circumference  of 
the  inscribed  and  circumscribed  circles. 

Ex.  6.  The  quadrilateral  ADBC  is  inscribed  in  a  circle.  The  diag- 
onals AB  and  DC  intersect  in  the  point  F.  Arc  AD  =  112°,  arc  A  C  = 
108°,  LAFC=  74°.  Find  all  the  other  angles  of  the  figure. 

Ex.  7.  Find  the  locus  of  the  center  of  a  circle  which  touches  two 
given  equal  circles. 

Ex.  8.  Find  the  area  of  a  triangle  whose  sides  are  1  m.,  17  dm., 
210  cm. 

Ex.  9.  The  line  joining  the  midpoints  of  two  radii  is  perpendicular 
to  the  line  bisecting  their  angle. 

Ex.  10.  If  a  quadrilateral  be  inscribed  in  a  circle  and  its  diag- 
onals drawn,  how  many  pairs  of  similar  triangles  are  formed  ? 

Ex.  11.  Prove  that  the  sum  of  the  exterior  angles  of  a  polygon 
(Art.  172)  equals  four  right  angles,  by  the  use  of  a  figure  formed  by 
drawing  lines  from  a  point  within  a  polygon  to  the  vertices  of  the 
polygon. 

(499) 


500  GEOMETRY.      APPENDIX 

Ex.  12.  In  a  circle  whose  radius  is  12  cm.,  find  the  length  of  the 
tangent  drawn  from  a  point  at  a  distance  240  mm.  from  the  center. 

Ex.  13.  If  two  sides  of  a  regular  pentagon  be  produced,  find  the 
angle  of  their  intersection. 

Ex.|14.  In  the  parallelogram  ABCD,  points  are  taken  on  the 
diagonals  such  that  AP=BQ=CR=DS.  Show  that  PQRS  is  a 
parallelogram. 

Ex.  15.  A  chord  6  in.  long  is  at  the  distance  4  in.  from  the  center 
of  a  circle.  Find  the  distance  from  the  center  of  a  chord  8  in.  long. 

Ex.  Ii6.  If  B  is  a  point  in  the  circumference  of  a  circle  whose* 
center  is \O,  PA  a  tangent  at  any  point  P,  meeting  OB  produced  at  A, 
and  PD  perpendicular  to  OB,  then  PB  bisects  the  angle  APD. 

Ex.  17.  Construct  a  parallelogram,  given  a  side,  an  angle,  and  a 
diagonal. 

Ex.  18.  Find  in  inches  the  sides  of  an  isosceles  right  triangle 
whose  area  is  1  sq.  yd. 

Ex.  19.    Given  the  line  a,  construct  qvi/2— 1)^ 

o 

Ex.  20.  If  two  lines  intersect  so  that  the  product  of  the  segments 
of  one  line  equals  the  product  of  the  segments  of  the  other,  a  cir- 
cumference may  be  passed  through  the  extremities  of  the  two  lines. 

Ex.  21.  Find  the  locus  of  the  vertices  of  all  triangles  on  a  given 
base  and  having  a  given  area. 

Ex.  22.    On  the  figure  p.  206,  prove  thatl^2+Z^=ZB2+JW2. 

Ex.  23.  The  area  of  a  rectangle  is  108  and  the  base  is  three  times 
th'i  altitude.  Find  the  dimensions. 

Ex.  24.  If,  on  the  sides  AC  and  BC  of  the  triangle  ABC,  the 
squares,  AD  and  BF,  are  constructed,  AF  and  DB  are  equal. 

Ex.  25.  If  the  angle  included  between  a  tangent  and  a  secant  is 
half  a  right  angle,  and  the  tangent  equals  the  radius,  the  secant 
passes  through  the  center  of  the  circle. 


REVIEW    EXERCISES    IN    PLANE    GEOMETRY        501 

Ex.  26.  The  sum  of  the  areas  of  two  circles  is  20  sq.  yds.,  and  the 
difference  of  their  areas  is  15  sq.  yds.  Find  their  radii. 

Ex.  27.  Construct  an  isosceles  trapezoid,  given  the  bases  and  a 
leg. 

Ex.  28.  Show  that,  if  the  alternate  sides  of  a  regular  pentagon 
be  produced  to  meet,  the  points  of  intersection  formed  are  the  vertices 
of  another  regular  pentagon. 

Ex.  29.  If  a  post  2  ft.  6  in.  high  casts  a  shadow  1  ft.  9  in.  long, 
how  tall  is  a  tree  which,  at  the  same  time,  casts  a  shadow  66  ft.  long  ? 

Ex.  30.  If  two  intersecting  chords  make  equal  angles  with  the 
diameter  through  their  point  of  intersection,  the  chords  are  equal. 

Ex.  31.  From  a  given  point  draw  a  secant  to  a  circle  so  that  the 
external  segment  is  half  the  secant. 

Ex.  32.  Find  the  locus  of  the  center  of  a  circle  which  touches  a 
given  circle  at  a  given  point. 

Ex.  33.  If  one  diagonal  of  a  quadrilateral  bisects  the  other 
diagonal,  the  first  diagonal  divides  the  quadrilateral  into  two  equi- 
valent triangles. 

Ex.  34.    In  a  given  square  inscribe  #  square  having  a  given  side. 

Ex.  35.  A  field  in  the  shape  of  an  equilateral  triangle  contains 
one  acre.  How  many  feet  does  one  side  contain  ? 

Ex.  36.  If  perpendiculars  are  drawn  to  a  given  line  from  the  ver- 
tices of  a  parallelogram,  the  sum  of  the  perpendiculars  from  two 
opposite  vertices  equals  the  sum  of  the  other  two  perpendiculars. 

Ex.  37.  Any  two  altitudes  of  a  triangle  are  reciprocally  propor- 
tional to  the  bases  on  which  they  stand. 

Ex.  38.  Construct  a  triangle  equivalent  to  a  given  triangle  and 
having  two  given  sides. 

Ex.  39.  The  apothem  of  a  regular  hexagon  is  20.  Find  the  area 
of  the  inscribed  and  circumscribed  circles. 

Ex.  40.  M  is  the  midpoint  of  the  hypotenuse  AB  of  e,  right  tri- 
angle ABC.  Prove  8  MC2=A£2+BC2-^AC2, 


502  GEOMETRY.      APPENDIX 

Ex.  41.  Transform  a  given  triangle  into  an  equivalent  right  tri- 
angle containing  a  given  acute  angle. 

Ex.  42.  The  area  of  a  square  inscribed  in  a  semicircle  is  to  the 
area  of  the  square  inscribed  in  the  circle  as  2  : 5. 

Ex.  43.  If,  on  a  diameter  of  the  circle  0,  OA  =  OB  and  AC  is  par- 
allel to  BD,  the  chord  CD  is  perpendicular  to  AC. 

Ex.  44.  Find  the  radius  of  a  circle  whose  area  is  equal  to  one- 
third  the  area  of  the  circle  whose  radius  is  7  in. 

Ex.  45.    State  and  prove  the  converse  of  Prop.  XXI,  Book  III. 

Ex.  46.  If,  in  a  given  trapezoid,  one  base  is  three  times  the  other 
base,  the  segments  of  each  diagonal  are  as  1  :  3. 

Ex.  47.  If  two  sides  of  a  triangle  are  6  and  12  and  the  angle 
included  by  them  is  60°,  fiad  the  length  of  the  other  side.  Also  find 
this  when  the  included  angle  is  45°;  also,  when  120°. 

Ex.  48.  How  many  sides  has  a  polygon  in  which  the  sum  of  the 
interior  angles  exceeds  the  sum  of  the  exterior  angles  by  540°? 

Ex.  49.  If  the  four  sides  of  a  quadrilateral  are  the  diameter  of  a 
circle,  the  two  tangents  at  its  extremities,  and  a  tangent  at  any  other 
point,  the  area  of  the  quadrilateral  equals  one-half  the  product  of  the 
diameter  by  the  side  opposite  it  in  the  quadrilateral. 

Ex.  50.  An  equilateral  triangle  and  a  regular  hexagon  have  the 
same  perimeter;  find  the  ratio  of  their  areas. 

Ex.  51.  To  a  circle  whose  radius  is  30  cm.  a  tangent  is  drawn 
from  a  point  21  dm.  from  the  center.  Find  the  length  of  the  tangent. 

Ex.  52.  If  two  opposite  sides  of  a  quadrilateral  are  equal,  and 
the  angles  which  they  make  with  a  third  side  are  equal,  the  quad- 
rilateral is  a  trapeaoid. 

Ex.  53.  If  two  circles  are  tangent  externally  and  two  parallel 
diameters  are  drawn,  one  in  each  circle,  a  pair  of  opposite  extremities 
of  the  two  diameters  and  the  point  of  contact  are  collinear. 

Ex.  54.  If,  in  the  triangle  ABC,  the  line  AD  is  perpendicular  to 
BD,  the  bisector  of  the  angle  B,  a  line  through  D  parallel  to  BC 
bisects  AC. 


REVIEW    EXERCISES    IN     PLANE     GEOMETRY        503 

Ex.  55.    Bisect  a  given  triangle  by  a  line  parallel  to  a  given  line. 

Ex.  56.  If  two  parallelograms  have  an  angle  of  one  equal  to  the 
supplement  of  an  angle  of  the  other,  their  areas  are  to  each  other  as 
the  products  of  the  sides  including  the  angles. 

Ex.  57.  The  sum  of  the  medians  of  a  triangle  is  less  than  the 
perimeter,  and  greater  than  half  the  perimeter. 

Ex.  58.  If  PARE  is  a  secant  to  a  circle  through  the  center  O,  PT 
a  tangent,  and  TE  perpendicular  to  PB,  then  PA  :  PR=PO  :  PB. 

Ex.  59.  Two  concentric  circles  have  radii  of  17  and  15.  Find  the 
length  of  the  chord  of  the  larger  which  is  tangent  to  the  smaller. 

Ex.  60.    On  the  figure,  p.  244, 

(a)  Find  two  pairs  of  similar  triangles; 

(1}  Find  two  dotted  lines  which  are  perpendicular  to  each  other; 

(c)  Discover  a  theorem  concerning  points,  not  connected  by  lines 
on  the  figure,  which  are  collinear; 

(d)  Discover  a  theorem  concerning  squares  on  given  lines. 

Ex.  61.  One  of  the  legs,  AC,  of  an  isosceles  triangle  is  produced 
through  the  vertex,  C,  to  the  point  F,  and  F  is  joined  with  D,  the  mid- 
point of  the  base  AB.  DF  intersects  BC  in  E,  Prove  that  CF  is 
greater  than  CE. 

Ex.  62.  The  line  of  centers  of  two  circles  intersects  their  common 
external  tangent  at  P.  PABCD  is  a  secant  intersecting  one  of  the 
two  circles  at  A  and  B  and  the  other  at  C  and  D.  Prove 


Ex.  63.  Trisect  a  given  parallelogram  by  lines  drawn  through  a 
given  vertex. 

Ex.  64.  Find  the  area  of  a  triangle  the  sides  of  which  are  the 
chord  of  an  arc  of  120°  in  a  circle  whose  radius  is  1  ;  the  chord  of  an 
arc  of  90°  in  a  circle  whose  radius  is  2  ;  and  the  chord  of  an  arc  of 
60°  in  a  circle  whose  radius  is  3. 

Ex.  65.  Construct  a  triangle,  given  the  median  to  one  side  and  a 
median  and  altitude  on  the  other  side. 

Ex.  66.  Two  circles  intersect  at  P  and  Q.  The  chord  CQ  is  tan- 
gent to  the  circle  QPB  at  Q.  APB  is  any  chord  through  P.  Prove 
that  AC  is  parallel  to  BQ. 


504  GEOMETRY.      APPENDIX 

Ex.  67.  In  the  triangle  ABC,  from  D,  the  midpoint  of  BC,  DE 
and  DF  are  drawn,  bisecting  the  angles  ADB  and  ADC,  and  meeting 
AB  at  E  and  AC  at  ^.  Prove  EF  \\  BC. 

Ex.  68.  Produce  the  side  BC  of  the  triangle  ABC  to  a  point  P,  so 
that  PBXPC=RA2. 

Ex.  69.  In  a  given  circle  inscribe  a  rectangle  similar  to  a  given 
rectangle. 

Ex.  70.  In  a  given  semicircle  inscribe  a  rectangle  similar  to  a 
given  rectangle. 

Ex.  71.  The  area  of  an  isosceles  trapezoid  is  140  sq.  ft.,  one  base 
is  26  ft.,  and  the  legs  make  an  angle  of  45°  with  the  other  base.  Find 
the  other  base. 

Ex.  72.  Cut  off  one-third  the  area  of  a  given  triangle  by  a  Hue 
perpendicular  to  one  side. 

Ex.  73.  Find  the  sides  of  a  triangle  whose  area  is  1  sq.  ft.,  if 
the  sides  are  in  the  ratio  2:3:4. 

Ex.  74.  Divide  a  given  line  into  two  parts  such  that  the  sum  of 
the  squares  of  the  two  parts  shall  be  a  minimum. 

Ex.  75.  If,  from  any  point  in  the  base  of  a  triangle,  lines  are 
drawn  parallel  to  the  sides,  find  the  locus  of  the  center  of  the  paral- 
lelogram so  formed. 

Ex.  76.  Three  sides  of  "a  quadrilateral  are  845,  613,  810,  and  the 
fourth  side  is  perpendicular  to  the  sicies  845  and  810.  Find  the  area. 

Ex.  77.  If  BP  bisects  the  angle  ABC,  and  DP 
bisects  the  angle  CDA,  prove  that  angle  P=i  sum 
of  angles  A  and  C. 

Ex.  78.    Two    circles    intersect   at   P  and    Q. 
Through  a  point  A  in  one  circumference  lines  APC 
and  AQD  are  drawn,  meeting  the  other  in  C  and  D.     Prove  the  tan- 
gent at  A  parallel  to  CD. 

Ex.  79.  In  a  given  triangle,  draw  a  line  parallel  to  the  base  and 
terminated  by  the  sides  so  that  it  shall  be  a  mean  proportional  be- 
tween thy  segments  of  one  side. 


KEYIEW    EXERCISES    IN    PLANE    GEOMETRY        505 

Ex.  80.  Find  the  angle  inscribed  in  a  semicircle  the  sum  of  whose 
Rides  is  a  maximum. 

Ex.  81.  The  bases  of  a  trapezoid  are  160  and  120,  and  the  alti- 
tude 140.  Find  the  dimensions  of  two  equivalent  trapezoids  into 
which  the  given  trapezoid  is  divided  by  a  line  parallel  to  the  base. 

Ex.  82.  If  the  diameter  of  a  given  circle  be  divided  into  any  two 
segments,  and  a  semicircumference  be  described  on  the  two  segments 
on  opposite  sides  of  the  diameter,  the  area  of  the  circle  will  be  di- 
vided by  the  semicircumferences  thus  drawn  into  two  parts  having 
the  same  ratio  as  the  segments  of  the  diameter. 

Ex.  83.  On  a  given  straight  line,  AB,  two  segments  of  circles  are 
drawn,  APB  and  AQB.  The  angles  QAP  and  QBP  are  bisected  by  lines 
meeting  in  R.  Prove  that  the  angle  JR  is  a  constant,  wherever  P  and 
Q  may  be  on  their  arcs. 

Ex.  84.  On  the  side  AB  of  the  triangle  ABC,  as  diameter,  a  cir- 
cle is  described.  EF  is  a  diameter  parallel  to  EC.  Show  that  EB 
bisects  the  angle  ABC. 

Ex.  85.  Construct  a  trapezoid,  given  the  bases,  one  diagonal,  and 
an  angle  included  by  the  diagonals. 

Ex.  86.  If,  through  any  point  in  the  common  chord  of  two  inter- 
secting circles,  two  chords  be  drawn,  one  in  each  circle,  through  the 
four  extremities  of  the  two  chords  a  circumference  may  be  passed. 

Ex.  87.  From  a  given  point  as  center  describe  a  circle  cutting  a 
given  straight  line  in  two  points,  so  that  the  product  of  the  distances 
of  the  points  from  a  given  point  in  the  line  may  equal  the  square  of  a 
given  line  segment. 

Ex.  88.  AB  is  any  chord  in  a  given  circle,  P  any  point  on  the 
circumference,  PM  is  perpendicular  to  AB  and  is  produced  to  meet 
the  circle  at  Q ;  AN  is  drawn  perpendicular  to  the  tangent  at  P. 
Prove  the  triangles  NAM  and  PAQ  similar. 

Ex.  89.  If  two  circles  ABCD  and  EBCF  intersect  in  B  and  C  and 
have  common  exterior  tangents  AE  and  DF  cut  by  EC  produced  at  G 
and  H,  then  GH2=£C2-\-'A£!2, 


506  GEOMETRY.      APPENDIX 

EXERCISES.     CROUP  87 

REVIEW    EXERCISES    IN    SOLID    GEOMETRY 

Ex.  1.  A  segment  of  a  straight  line  oblique  to  a  plane  is  greater 
than  its  projection  on  the  plane. 

Ex.  2.  Two  tetrahedrons  are  similar  if  a  dihedral  angle  of  one 
equals  a  dihedral  angle  of  the  other,  and  the  faces  forming  these 
dihedral  angles  are  similar  each  to  each. 

Ex.  3.  A  plane  and  a  straight  line,  both  of  which  are  parallel  to 
the  same  line,  are  parallel  to  each  other. 

Ex.  4.  If  the  diagonal  of  one  face  of  a  cube  is  10  inches,  find  the 
volume  of  the  cube. 

Ex.  5.  Construct  a  spherical  triangle  on  a  given  sphere,  given  the 
poles  of  the  sides  of  the  triangle. 

Ex.  6.    Given  AB  _L  MN, 

AE  and  BF  _L  ME; 
prove  EF  J.  PM. 

Ex.  7.    The  diagonals  of  a  rectangular  paral- 
lelepiped are  equal. 

Ex.  8.    What  portion  of  the  surface  of  a  sphere 
is  a  triangle  each  of  whose  angles  is  140°? 

Ex.  9.  Through  a  given  point  pass  a  plane  parallel  to  two  given 
straight  lines. 

Ex.  10.  Show  that  the  lateral  area  of  a  cylinder  of  revolution  is 
equivalent  to  a  circle  whose  radius  is  a  mean  proportional  between 
the  altitude  of  the  cylinder  and  the  diameter  of  its  base. 

Ex.  11.  The  volumes  of  polyhedrons  circumscribed  about  equal 
spheres  are  to  each  other  as  the  surfaces  of  the  polyhedrons. 

Ex.  12.  Find  /Sand  T  of  a  regular  square  pyramid  an  edge  of 
whose  base  is  14  dm.,  and  whose  lateral  edge  is  250  cm. 

Ex.  13.  If  two  lines  are  parallel  and  a  plane  be  passed  through  each 
line,  the  intersection  of  these  plane?  is  parallel  to  the  given  lines. 


REVIEW    EXEECISES     IN    SOLID    GEOMETRY        507 


If 


Ex.  14.    Given  PH  J_  plane  AD, 

/.PEH=  /.PFH; 
prove  Z  PEF=  Z  PFE. 

Ex.  15.    If  a  plane   be   passed 
through    the    midpoints    of    three    <?Z 
edges    of    a    parallelepiped    which 

meet    at   a   vertex,   the   pyramid  thus  formed   is  what  part  of   the 
parallelepiped  ? 

Ex.  16.  Find  a  point  in  a  plane  such  that  the  sum  of  its  distances 
from  two  given  points  on  the  same  side  of  the  plane  is  a  minimum. 

Ex.  17.  Given  the  points  A,  B,  C,  D  in  a  plane  and  P  a  point 
outside  the  plane,  AB  perpendicular  to  the  plane  PBD,  and  AC  per- 
pendicular to  the  plane  PCD;  prove  that  PD  is  perpendicular  to 
the  plane  ABCD. 

Ex.  18.  In  a  sphere  whoso  radius  is  5,  find  the  area  of  a  zone  the 
radii  of  whose  upper  and  lower  bases  are  3  and  4. 

Ex.  19.  Two  cylinders  of  revolution  have  equal  lateral  areas. 
Show  that  their  volumes  are  as  R  :  R' . 

Ex.  20.  The  midpoints  of  two  opposite  sides  of  a  quadrilateral  in 
space,  and  the  midpoints  of  its  diagonals,  are  the  vertices  of  a 
parallelogram. 

Ex.  21.  How  many  feet  of  two-inch  plank  are  necessary  to  con- 
struct a  box  twice  as  wide  as  deep  and  twice  as  long  as  wide  (on  the 
inside),  and  to  contain  216  cu.  ft.? 

Ex.  22o  If  two  spheres  with  radii  R  and  r  are  concentric,  find  the 
area  of  the  section  of  the  larger  sphere  made  by  a  plane  tangent  to 
the  smaller  sphere. 

Ex.  23.  In  the  frustum  of  a  regular  square  pyramid,  the  edges 
of  the  bases  are  denoted  by  &i  and  &•_>  and  the  altitude  by  H;  prove 


that  i=li(61—  62)2-f4JH"2. 

Ex.  24.     If  the  opposite  sides  of  a  spherical  quadrilateral  are  equal 
the  opposite  angles  are  equal. 


508  GEOMETRY.      APPENDIX 

Ex.  25.  Obtain  the  simplest  formula  for  the  lateral  surface  of  a 
truncated  triangular  right  prism,  each  edge  of  whose  base  is  a,  and 
whose  lateral  edges  are  p,  q,  and  r. 

Ex.  26.  The  area  of  a  zone  of  one  base  is  a  mean  proportional 
between  the  remaining  surface  of  the  sphere  and  its  entire  surface. 
Find  the  altitude  of  the  zone. 

Ex.  27.  The  lateral  edges  of  two  similar  frusta  are  as  1  :  a.  How 
do  their  areas  compare  ?  Their  volumes  ? 

Ex.  28.  Construct  a  spherical  surface  with  a  given  radius,  r,  which 
shall  be  tangent  to  a  given  plane,  and  to  a  given  sphere,  and  also  pass 
through  a  given  point. 

Ex.  29.  The  volume  of  a  right  circular  cylinder  equals  the  area 
of  the  generating  rectangle  multiplied  by  the  circumference  generated 
by  the  point  of  intersection  of  its  diagonals. 

Ex.  30.  On  a  sphere  whose  radius  is  8i  inches,  find  the  area  of  a 
zone  generated  by  a  pair  of  compasses  whose  points  are  5  inches 
apart. 

Ex.  31.  The  perpendicular  to  a  given  plane  from  the  point  where 
the  altitudes  of  a  regular  tetrahedron  intersect  equals  one -fourth  the 
sum  of  the  perpendiculars  from  the  vertices  of  the  tetrahedron  to  the 
same  plane. 

Ex.  32.  Two  trihedral  angles  are  equal  or  symmetrical  if  their 
corresponding  dihedral  angles  are  equal. 

Ex.  33.  On  a  sphere  whose  radius  is  a,  a  zone  has  equal  bases 
and  the  sum  of  the  bases  equals  the  area  of  the  zone.  Find  the  alti- 
tude of  the  zone. 

Ex.  34.  A  plane  which  bisects  two  opposite  edges  of  a  tetrahedron 
bisects  the  volume  of  the  tetrahedron. 

Ex.  35.  Find  the  locus  of  all  points  in  space  which  have  their 
distances  from  two  given  parallel  lines  in  a  given  ratio. 

Ex.  36.  If  a,  b,  c  are  the  sides  of  a  spherical  triangle,  a',  b',  cf 
the  sides  of  its  polar  triangle,  and  a>&>c,  then  a'<b'<d '. 

Ex.  37.  A  cone  of  revolution  has  a  lateral  area  of  4  sq.  yd.  and 
an  altitude  of  2  ft.  How  much  of  the  altitude  must  be  cut  off  by  a 
plane  parallel  to  the  base,  in  order  to  leave  a  frustum  whose  lateral 
area  is  2  sq.  ft.  ? 


REVIEW    EXERCISES     IN     SOLID     GEOMETRY        509 

Ex.  38.  The  total  area  of  an  equilateral  cone  is  to  the  area  of  the 
inscribed  sphere  as  9  :  4. 

Ex.  39.  Construct  a  sphere  of  given  radius,  r,  whose  surface  shall 
be  tangent  to  three  given  spheres. 

Ex.  40.  The  volume  of  the  frustum  of  an  equilateral  cone  is  300 
cu.  in.  and  its  altitude  is  20  in.  Show  how  to  find  the  radii  of  the  bases. 

Ex.  41.  On  each  base  of  a  cylinder  of  revolution  a  cone  is  placed, 
with  its  vertex  at  the  center  of  the  opposite  base.  Find  the  radius  of 
the  circle  of  intersection  of  the  two  conical  surfaces. 

Ex.  42.  The  volume  of  a  frustum  of  a  cone  of  revolution  equals 
the  sum  of  a  cylinder  and  a  cone  of  the  same  altitude  as  the  frustum, 
and  with  radii  which  are  respectively  the  half  sum  and  the  half  differ- 
ence of  the  radii  of  the  frustum. 

Ex.  43.  A  square  whose  side  is  a  revolves  about  a  line  through 
one  of  its  vertices  and  parallel  to  a  diagonal,  as  axis;  find  the  surface 
and  volume  generated. 

Ex.  44.  If  a  cone  of  revolution  roll  on  another  fixed  cone  of  revo- 
lution so  that  their  vertices  coincide,  find  the  kind  of  surface  gen- 
erated by  the  axis  of  the  rolling  cone. 

Ex.  45.  An  equilateral  triangle  whose  side  is  a  revolves  about  an 
altitude  as  an  axis;  find  the  surface  and  volume  generated  by  the 
inscribed  circle,  and  also  by  the  circumscribed  circle. 

Ex.  46.  Find  the  locus  of  the  center  of  a  sphere  which  is  tan- 
gent to  three  given  planes. 

Ex.  47.  If  an  equilateral  triangle  whose  side  is  a  be  rotated  about 
a  line  through  one  vertex  and  parallel  to  the  opposite  side,  as  an  axis, 
find  the  surface  and  volume  generated. 

Ex.  48.  What  other  formulas  of  solid  geometry  may  be  regarded 
as  special  cases  of  the  formula  for  the  volume  of  a  prismatoid  ? 

Ex.  49.  Through  a  given  point  pass  a  plane  which  shall  bisect  the 
volume  of  a  given  tetrahedron. 

Ex.  50.  In  an  equilateral  cone  and  a  cone  whose  opposite  ele- 
ments are  perpendicular  at  the  vertex,  show  that  the  ratio  of  tli§ 
Vertical  solid  angles  is  as  2— j/3  :  2 — j/2. 


APPLICATIONS  OF  SOLID   GEOMETRY  TO  MECHANICS 
AND  ENGINEERING 


EXERCISES.       GROUP    93 

(BOOKS   VI   AND   VII) 

1.  A  carpenter  tests  the  flatness  of  a  surface  by  applying  a  e  traight 
edge  to  the  surface  in  various  directions.     How  does  a  plasterer  test 
the  flatness  of  a  wall  surface?     What  geometric  principle  is  used  by 
these  mechanics? 

2.  Explain  why  an  object  with  three   legs,  as    a  stool  or   tripod, 
always  rests  firmly  on  the  floor  while  an  object  with  four  legs,  as  a  table, 
does  not  always  rest  so.     Why  do  we  ever  use  four-legged  pieces  of 

30'  furniture? 

3.  How    can    a    carpenter 
get  a  corner  post  of  a  house 
in  a  vertical  position  by  use  of 
a   carpenter's    square?     What 
geometrical  principle  does  he 
use? 

4.  The  diagram  is  the  plan 
of  a  hip  roof.     The  slope  of 
each  face  of  the  roof  is  30°. 
Find  the  length  of  a  hip  rafter 
as  AB. 


20 


1      2 

[SUG. 


C 


Draw  a  triangle  DBC  representing  a  section  of  the  roof  at 

20 

DBC  on  the  plan.     Hence  it  may  be  shown  that  BC  =  —  V3.     In 

o 

like  manner  by  taking  a  section  through 
BF,  it  is  found  that  AC  =  10.  Hence  in 
The  triangle  ABC,  AB  may  be  found.] 

6.   Find   the  area  of  the  entire  roof 
represented  in  Ex.  4. 

6.    Make  drawings  showing  at-  what 

angle  the  two  ends  of  a  rafter  like  BC  in  Ex.  4  must  be  cut. 

510 


10' 


PRACTICAL  APPLICATIONS  511 

7.  Make  drawings  showing  at  what  angle  a  jack  rafter  like  12  in 
Ex.  4  must  be  cut. 

[Sue.  To  determine  how  the  end  1  of  the  jack  rafter  must  be  cut 
use  the  principle  that  two  intersecting  straight  lines  determine  a  plane 
(Art.  501).  The  cutting  plane  at  1  'must  make  an  angle  at  the  side 
of  the  jack  rafter  equal  to  angle  CBH,  and  on  the  top  of  the  jack  rafter 
equal  to  angle  ABC.] 

8.  What  is  a  gambrel  roof?     Make  up  a  set  of  examples  concerning 
a  gambrel  roof  similar  to  Exs.  4-7. 

9.  By  use  of  Art.  645,  show  that  a  page  of  this  book  held  at  twice 
the  distance  of  another  page  from  the  same  lamp  receives  one  fourth 
the  light  the  first  page  receives. 

10.  The  supporting  power  of  a  wooden  beam  varies  directly  as  the 
area  of  the  cross  section  times  the  height  of  the  beam  and  inversely  as 
the  length  of  the  beam.  Compare  the  supporting  power  of  a  beam 
12  ft.  long,  3  in.  wide,  and  6  in.  high  with  that  of  a  beam  18  ft.  long,  4  in. 
wide,  and  10  in.  high.  Also  compare  the  volumes  of  the  two  beams. 


EXERCISES.       GROUP    94 

(BOOKS  VIII  AND   IX) 

1.  A  hollow  cylinder  whose  inside  diameter  is  6  in.  is  partly  filled 
with  water.     An  irregularly  shaped  piece  of  ore  when  placed  in  the 
water  causes  the  top  surface  of  the  water  to  rise  3.4  in.  in  the  cylinder. 
Find  the  volume  of  the  ore. 

2.  What  is  a  tubular  boiler?     What  is  the  advantage  in  using  a 
tubular  boiler  as  compared  with  a  plain  cylindrical  boiler?     If  a  tubular 
boiler  is  18  ft.  long  and  contains  32  tubes  each  3  in.  in  diameter,  how 
much  more  heating  surface  has  it  than  a  plain  cylindrical  boiler  of  the 
same  length  and  36  in.  in  diameter?    (Indicate  both  the  long  method  and 
the  short  method  of  making  this  computation  and  use  the  short  method.) 

3.  If  a  bridge  is   to   have   its   linear   dimensions    1000   times   as 
great  as  those  of  a  given  model,  the  bridge  will  be  how  many  times 
as  heavy  as  the  model? 

Why,  then,  may  a  bridge  be  planned  so  that  in  the  model  it  will  sup- 
port relatively  heavy  weights,  yet  when  constructed  according  to  the 
model,  falls  to  pieces  of  its  own  weight? 

Show  that  this  principle  applies  to  other  constructions,  such  as 
buildings,  machines,  etc.,  as  well  as  to  bridges. 


512  GEOMETRY 

4.  Work  again  Exs.  23-25,  27,  p.  473. 

5.  Make  and  work  for  yourself  an  example  similar  to  Ex.  24,  p.  473. 

6.  Sound  spreads  from  a  center  in  the  form  of  the  surface  of  an 
expanding  sphere.     At  the  distance  of  10  yd.  from  the  source,  how  will 
the  surface  of  this  sphere  compare  with  its  surface  as  it  was  at  1  yd.? 
How,  then,  does  the  intensity  of  sound  at  10  yd.  from  the  source  com- 
pare with  its  intensity  at  a  distance  of  1  yd.? 

Does  this  law  apply  to  all  forces  which  radiate  or  act  from  a  center 
as  to  light,  heat,  magnetism,  and  gravitation?  Why  is  it  called  the 
law  of  inverse  squares? 

7.  If  a  body  be  placed  within  a  spherical  shell,  the  attractive  forces 
exerted  upon  the  body  by  different  parts  of  the  shell  will  balance  or 
cancel  each  other.     Hence  a  body  inside  the  earth,  as  at  the  foot  of  a 
mine,  is  attracted  effectively  only  by  the  sphere  of  matter  whose  radius 
is  the  distance  from  the  center  of  the  earth  to  the  given  body.   Hence, 
prove  that  the  weight  of  a  body  below  the  surface  of  the  earth  varies 
as  the  distance  of  the  body  from  the  center  of  the  earth. 

[Sua.  If  W  denote  the  weight  of  the  body  at  the  surface,  and  w  its 
weight  when  below,  R  the  radius  of  the  earth,  and  r  the  distance  of  the 

body  from  the  center  when  below 
the  surface,  show  that 


8.  The  light  of  the  sun  falling 
on  a  smaller  sphere,  as  on  the  earth  or  the  moon,  causes  that  body  to 
cast  a  conical  shadow.  Denoting  the  radius  of  the  sun  by  R,  the 
radius  of  the  smaller  sphere  by  r,  the  distance  between  the  two  spheres 

dr 

by  d,  and  the  length  of  the  shadow  by  Z,  show  that  I  =  — • 

ri  —  r. 

Find  I  when  d  =  92,800,000  mi.,  R  =  .433,000  mi.,  r  =  4000  mi. 

9.  If  in  a  lunar  eclipse  the  moon's  center  should  pass  through  the 
axis  of  the  conical  shadow,  and  the  moon  is  traveling  at  the  rate  of 
2100  mi.  an  hour,  how  long  would  the  total  eclipse  of  the  moon  last? 

flow  long  if  the  moon's  center  passed  through  the  earth's  conical 
shadow  at  a  distance  of  1000  mi.  from  the  axis  of  the  cone? 


PRACTICAL  APPLICATIONS  513 

10.  If  the  moon's  diameter  is  2160  mi.,  find  the  length  of  the  moon's 
shadow  as  caused  by  sunlight. 

11.  If  the  distance  of  the  moon  from  the  earth's  center  varies  from 
221,600  mi.  to  252,970,  show  how  this  explains  why  some  eclipses  of  the 
sun  are  total  and  others  annular. 

Why,  also,  at  a  given  point  on  the  earth's  surface  is  an  eclipse  of  the 
sun  a  so  much  rarer  sight  than  an  eclipse  of  the  moon?  Why,  also,  is  its 
duration  so  much  briefer? 

12.  Prove  that  the  latitude  of  a  place   on    the 
earth's    surface    equals    the   elevation   of   the   pole 
(that  is,  on  the  diagram,  prove  /_  QEA  =  /_PAO). 

^       Z  13.    Given   the  sun's   declina- 

tion (i.e.  distance  north  or  south 
of  the  celestial    equator),    show 
how  to  determine    the    latitude 
H  •*    of  a  place  by  measuring  the  zenith  distance  of  the 

sun.     Also  by  measuring  the  altitude  of  the  sun  above  the  horizon. 

14.  How  was  Peary  aided  by  the  principles  of  Exs.  26  and  27  in 
determining  whether  he  had  arrived  at  the  North  Pole? 

16.  If  on  April  6  (the  day  of  the  year  on  which  Peary  was  at  the 
North  Pole)  the  sun  was  6°  7'  north  of  the  celestial  equator,  how  high 
above  the  horizon  should  the  sun  have  been  as  observed  by  Peary? 
At  what  hour  of  the  day  was  this? 

16.  What  is  the  sextant?     Explain  and  prove  the  principle  of  the 

sextant. 

17.  Explain  and  prove  the  principle  of  the  angle-meter. 


FORMULAS  OF  PLANE  GEOMETRY 

SYMBOLS 

a,  6,  c= sides  of  triangle  ABC.  K= radius  of  a  circle. 

s=$  (a-f  6  +  c).  D= diameter  of  a  circle. 

7ic=altitude  on  side  c.  C=  circumference  of  a  circle. 

me= median  on  side  c.  r= radius   of    an    inscribed 

/c=bisector  of  angle  opposite  circle. 

side  c.  T=^7-approx.  (or  3.1416—). 

I  and  m  =  line  segments.  .fir=area. 

P=perimeter.  &=base  of  a  triangle. 

>S»=side  of  a  regular  polygon  li  —  altitude  of  a  triangle. 
of  n  sides.                        61  and  &2=bases  of  a  trapezoid. 

LENGTHS    OF    LINES 

1.  In  a  right  triangle,  C  being  the  right  angle, 

c2=a2  +  62.  Art.  346. 

2.  In  a  right  triangle,  I  and  m  being  the  projections  of  a  and  b 
on  c,  and  h,  the  altitude  on  c, 

a2  =  ZXc,WmXc.  Art.  342. 

3.  In  an  oblique  triangle,  tn  being  the  projection  of  &  on  c, 

if  a  is  opposite  an  acute  Z  ,  a'2  =  62  -j-  c2  —  2  cX  m>     Art.  349. 
if  a  is  opposite  an  obtuse  L  ,  a2  =  &2  +  c2  +  2  c  X  w.     Art.  350. 


4.    fcc=tl/s(s  —  a)  (s  —  6)  (s  —  c).  Art.  393. 


5.    wc=4-|/2  (a2  +  62)  —  c2.  Art.  353. 


6.  fc=— — f/a&s  (s  —  c).  Art.  363. 

7.  If  I  and  m  are  the  segments  of  c  made  by  the  bisector  of  the 
angle  opposite,  a:b  =  l:m.  Arts.  332,  336. 

(514} 


FORMULAS  OF  PLANE  GEOMETKY  515 

8.  If  I  and  m  are  the  segments  of  a  line,  a,  divided  in  extreme  and 
mean  ratio,  and  I  >  m,                 a  :l=l  :m.  Art.  370. 

(P:Pf =«:«'.  Art.  341. 

9.  In  similar  polygons,   (  ^  ;  ^  ;  &/  ^   ^ 

10.  In  circles,  C  :  C'  =  R  :  R' ;  also  C  :  C'=>D  :  D1 .  Art.  442. 

11.  C=2  irR,  or  C=irD.  Art.  444. 

12.  An  arc  =  eB*r*™*leX*B.  Art.  445. 

J-oU 

13.  In  inscribed  regular  polygons, 


S2n=V  R  ( 2  R  — 1/4  R>—  Sn2) .  Art.  467 . 

AREAS    OF    PLANE    FIGURES 

1.    In  a  triangle,  K=l  Ih.  Art.  389. 


2.  In  a  triangle,  K=\/s  (s—a)  (s  —  b)   (s  —  c).  Art.  393. 

3.  In  an  equilateral  triangle,  K=C?^.  Ex.  4,  p.  257. 

4.  In  a  parallelogram,  K=by^li.  Art.  385. 

5.  .  In  a  trapezoid,  K=i  h  (61  +  &2).  Art.  394. 

6.  In  a  regular  polygon,  K=$  rXP.  Art.  446. 

7.  In  a  circle,  K^I?  or  JT=i  vD\  Art.  442. 

8.  In  a  sector  of  a  circle,  Z"=i  i?X  arc,  Art.  453. 


9.  In  a  segment  of  a  circle,  jK^=  sector  ±  A  formed  by  the  chord 
and  radii  of  the  segment. 

10.  In  a  circular   ring,  E=Tr  (R2  —  Rn).  Art.  449. 

11.  In  any  two  similar  plane  figures, 

K:K'=a*  :a/2;  Art.  399, 

also  a  :  a'  =  /K  :  /!7.  Art.  314. 


12.    In  two  circles,  K  :  K'=R2  :  .R/2  =  Z>2  :  Z>/2=C2  :  C/2;        Art.  452. 
also  R  :  R'=D  :  D>  =  C  :  C'=/K  :   /tf'.     Art.  314. 


FOEMULAS  OF  SOLID  GEOMETRY 

SYMBOLS 

B,  &=areas  of  the  lower  and  up-  P=perimeter  of  right  section. 

per  bases  of  a  frustum.  P,  p  =  perimeters  of  lower  and  up- 

l£=lateral  edge  (or  element);  per  bases  of  a  frustum. 

or  r,  rf  =  radii  of  bases. 

=  spherical  excess.  £=  area  of  lateral  surface;  or 

H=  altitude.  =area  of  surface  of  sphere, 

I,  b,  7&  =  length,  breadth,  height.  etc. 

i=slant  height.  T=area  of  total  surface. 

<Jf=area  of  midsection.  V—  volume. 

FORMULAS    FOR    AREAS 

1.  In  a  prism,  S=EXP>  Art.  608. 

2.  In  a  regular  pyramid,  S=i  LX?.  Art.  641. 

3.  In  a  frustum  of  a  regular  pyramid,  S=i(P-\-p)  L.  Art.  643. 

4.  In  a  cylinder  of  revolution,  S=2  ^BH.  Art.  697. 

T=27rB(R  +  H).  Art.  697. 

5.  In  a  cone  of  revolution,  S=TRL.  Art.  721. 

T=TTK  (L  +  E).  Art.  721. 

6.  In  a  frustum  of  a  cone  of  revolution,  S=^L  (B-\-  r).  Art.  727. 

7.  In  a  sphere,  S=4  irg2,  Or  S=irIP.  Art.  810. 

8.  In  a  zone,  £=2  ^EH.  Art.  813. 


7J-J?2  A 

9.    In  a  lune,  8=-~~  .  Art.  817. 


10.    In  a  spherical  triangle,  S^-.  Art.  822. 

loO 


11.    In  a  spherical  polygon,  s=-~  Art.  824. 

180 

(516) 


VJi-'  '       •  *  2 


FOKMULAS     OF     SOLID     GEOMETRY  517 

FORMULAS    FOR    VOLUMES 

1.  In  a  prism,  V=B^H.  Art.  628. 

2.  In  a  parallelepiped,  F=lX^Xh.  Art.  626. 

3.  In  a  pyramid,  F=i  BX  H.  Art.  651. 

4.  In  a  frustum  of  a  pyramid,  F=i  H  (B  +  1  -\-\/  Bb).  Art.  656. 

5.  In  a  prismatoid,  V=\  H  (B  +  6  +4  M).  Art.  663. 

6.  In  a  cylinder,  V=BXH-  Art.  698. 

7.  In  a  cylinder  of  revolution,  V=irR'iH.  Art.  699. 

8.  In  a  cone,  F=i  BXH.  Art.  722. 

9.  In  a  circular  cone,  F=t  TJ22^.  Art.  723. 

10.  In  a  frustum  of  a  cone,  V=%  H  (B-\-l  -\-\/  Bb).  Art.  729. 

11.  In  a  frustum  of  a  cone  of  revolution, 

V=\TTR(  U2  +  r2  +  Rr)  .  Art  .  730  . 

12.  In  a  sphere,  F=£  ^.B3,  or  F=|7rDs.  Art.  832. 

13.  In  a  spherical  sector,  F=t  ^-B2^.  Art.  836. 

14.  In  a  spherical  segment  of  two  bases, 


F=i  (Trr2  +  7T/2)  Jff  +  i  7T#3.  Art.  837. 

15.    In  a  spherical  segment  of  one  base,  F=TJH2  (R  —  ^H). 

Art.  838. 

CONSTANTS 

1  aere  =  43,560  sq.  ft.  ^2  =  1.2599  + 

1  bushel  =  2150.42  cu.  in.  ^3  =  1.4422  + 

1  gallon  =  231  cu.  in,  ^=-3183  + 

l/2  =  1.4142+  v/7r=1.7725- 

!/3  =  1.7321  -  ^=0.5642  + 

GG 


513  GEOMETRY.      APPENDIX 

SUMMARY    OF    METRIC    SYSTEM 

TABLE    FOR   LEHGTH 

10  millimeters  (mm.)  =1  centimeter  (cm.) 

10  cm.  =1  decimeter  (dm.) 

10  dm.  =1  meter  (m.). 

10  m.  =1  Dekameter  (Dm.) 

10  Dm.  =1  Hektometer  (Hm.) 

10  Hm.  =1  Kilometer  (Km.) 

10  Km.  =  1  Myriameter  (Mm.) 

Similar  tables  are  used  for  the  unit  of  weight,  the  gram ;  for  the 
unit  of  capacity,  the  liter  ;  for  the  unit  of  land  measure,  the  are;  and 
for  the  unit  of  wood  measure,  the  stere. 

TABLE   FOR   SQUARE   MEASURE 

100  sq.  mm.  =  l  sq.  cm. 

100  sq.  cm.  =1  sq.  dm.,  etc. 

TABLE  FOR  CUBIC  MEASURE 

1000  cu.  mm.  =  l  cu.  cm. 
1000  cu.  cm.  =1  cu.  dm.,  etc. 

A  liter  =1  cu.  dm. 

A  gram  —  weight  of  1  cu.  cm.  of  water  at  39.2°  Fahrenheit. 

An  are  =100  sq.  m. 

A  stere  =1  cu.  m. 

EQUIVALENTS 

1  meter       =39.37  inches. 
1  liter         =1.057  liquid  quarts, 
or  .9581  dry  quarts. 
1  kilogram  =  2. 2046  Ibs.  av. 
1  hektare    =2.471  acres. 
1  sq.  m.        —  1550—  sq.  in. 


INDEX   OF   DEFINITIONS   AND   FORMULAS 


PAGE 

Altitude  of  cone  .      .      .      .412 

of  cylinder 403 

of  frustum  of  cone      .      .414 
of  frustum  of  pyramid     .    378 

of  prism 3d 

of  pyramid         ....   377 
of  spherical  segment    .      .   402 

of  zone 452 

Angle,  dihedral  .  .  .  .337 
formed  by  two  curves  .  433 
of  lune  ......  452 

polyhedral 349 

spherical 433 

tetrahedral 349 

trihedral 349 

Angles  of  spherical  polygon  438 

Appolonius 493 

Arabs 490 

Archimedes       .      .      .        497,498 

Aryabhatta 497 

August,  E.   F 498 

Axis  of  circle  of  sphere  .  427 
of  circular  cone  .  .  .413 
of  regular  pyramid  .  .  378 
of  sphere 427 

Babylonians 497 

Base  of  cone  .  412 

of  pyramid 377 

of  spherical  pyramid  .      .461 
of  spherical  sector  .      .      .461 


PAGE 

Bases  of  cylinder  .  .  .403 
of  frustum  of  cone  .  .414 
of  frustum  of  pyramid  .  378 

of  prism 361 

of  spherical  segment    .      .   462 

of  zone 452 

Bodies,  The  Three  Round     .   492 

Center  of  sphere  ....  425 

Circle 496 

great 427 

small .427 

Cone 412 

altitude  of 412 

axis  of 413 

base  of 412 

circular         413 

circular,  formula  for  vol- 
ume of 418 

circular,  formulas  for  lat- 
eral, a  total  area  of     .417 

elements  of 412 

lateral  surface  of  .      .      .412 

oblique  circular      .      .      .413 

of    revolution    .      .      .      .413 

right  circular    .      .      .      .413 

vertex  of      .....  412 

Cones,  similar       .      .      .      .413 

Conical    surface    .      .      .      .412 

directrix  of        .      .      .      .412 

element  of          ....  412 


519 


520 


INDEX  OF  DEFINITIONS  AND  FORMULAS 


PAGE 

Conical    surface,    generatrix 

of 412 

nappes    of 412 

vertex  of 412 

Constants 513 

Continuity,  principle  of       .  485 

Cube 363 

Curved  spaces       ....  488 

Cylinder 403 

altitude  of          ....  403 

bases  of 403 

circular 404 

circular,  properties  of       .  405 

elements  of 403 

lateral  surface  of  ...  403 

oblique 404 

of  revolution     ....  404 
of  revolution,  formulas  for 
lateral   and   total   areas 

of 409 

right 404 

right  circular    ....  404 

right  section  of      ...  405 

section  of 405 

Cylinders       of       revolution, 

similar 404 

Cylindrical  surface     .      .      .  403 

directrix  of 403 

element  of 403 

generatrix   of    ....  403 

Degree,  spherical  ....   452 
Determined  plane       ,       319,320- 
Diagonal    of    polyhedron      .   360 
Diameter  of  sphere     .      .      .   425 
Dihedral  angle     .      .      .      .337 

edge  of 337 

faces   of 337 

plane  angle  of  .      .      .      .338 
right 337 


PAGE 

Dihedral  angles,  adjacent  .  337 

equal 337 

vertical  337 

Directrix  of  conical  surface  412 
of  cylindrical  surface  .  403 

Distance  from  point  to  plane  328 
on  surface  of  sphere  .  .  428 
polar,  of  great  circle  .  .  429 
polar,  of  small  circle  .  .  429 

Dodecahedron        .      .      .      .360 

Duality,  principle  of      .      .  486 

Edge  of  dihedral  angle  .  .  337 
Edges  of  polyhedral  angle  .  349 

of  polyhedron  ....  360 
Egyptians .  490,  491,  492,494,  497 
Elements  of  conical  surface  412 

of  cylindrical  surface  .  403 
Equivalent  solids  .  .  .  363 

Euclid 491 

Eudoxus  .  .  .  493,496,498 

Eulor 497 

European 490 

Faces  of  dihedral  angle  .      .   337 

of  polyhedral  angle     .      .   349 

of  polyhedron    ....   360 

Figures,  rectilinear     .      .      .   494 

Formulas  of  plane  geometry 

for  lengths  of  lines  514,  515 
for    plane    geometry,    for 

areas  of  plane  figures    .    515 
of     solid     geometry,     for 

areas 516 

of  solid  geometry,  for  vol- 
umes     517 

Foot  of  line 320 

Frustum    of    cone      .      .      .414 

altitude  of 414 

bases  of  .  .  414 


INDEX  OF  DEFINITIONS  AND  FORMULAS 


521 


PAGE 

Frustum  of  cone,  formula  for 

lateral  area  of  ...  420 
formula  for  volume  of  .  422 
lateral  surface  of  .  .  .414 
slant  height  of  .  .  .414 

Frustum  of  pyramid        .      .378 
altitude  of   .....   378 

bases  of  ......    378 

slant  height  of       .      .      .378 

Generatrix  of  conical  surface  412 
of  cylindrical  surface        .    403 

Geometry,  epochs  in  develop- 

ment of     .      .      .      .      .490 

history  of     ...        490-498 
modern    ......   485 

Non-Euclidean  .      .        488,496 
origin   of      .....   490 

projecti  ve     .....   485 

solid         ......   319 


Gerard 
Greeks  . 


498 
490,492,493,494 


Hero      .......   496 

Hexahedron      .....   360 

Hindoos       .      .      .     490,492,497 
Hippasius  ......   498 

Hippocrates  .  .  491,492,496 
History  of  geometry  .  490-498 
Homology,  principle  of  .  487 
Hyperspace  .....  488 

Icosahedron  .....  360 
Inclination  of  line  to  plane  347 


Jews 


497 


Lateral  area  of  frustum  of 

pyramid,  formula  for     .   379 


Lateral   area  of   frustum 
prism 

of  pyramid  .... 
Lateral  edges  of  -prism     . 

of  pyramid  .... 
Lateral  faces  of  prism    . 

of  pyramid  .... 
Lateral   surface   of   cone 

of  cylinder   .... 

of  frustum  of  cone 
Line  parallel  to  plane 

perpendicular  to  plane 
Lobatchewsky 
Logical  methods   . 
Lune 

angle  of 

formula    for    area    ofj 
spherical   degrees 

formula    for    area    of, 
square  units  of  area 


PAGE 

of 

.  361 

.  377 

.  361 

.  377 

.  361 

.  377 

.  412 

.  403 

.  414 

.  321 

.  320 

.  498 

.  492 

.  452 

.  452 

in 

.  457 

in 

.  457 


Mechanical  methods  .      .      .   493 

Menelaus 498 

Methods,  logical   .      .      .      .492 

mechanical 493 

rhetorical  .  .  .  .  .491 
Metric  system,  summary  of  514 
Modern  geometry  .  .  .485 

Nappes  of  cone  .  .  .  .412 
Negative  quantities  .  485,  486 
Non-Euclidean  geometry 

488,  498 


Octahedron 
Origin  of  geometry 

.      .      .   360 

.      .      .   490 

Parallelepiped 
right 

.      .      .   362 
362 

rectangular 

.      .      .  363 

522 


INDEX  OF  DEFINITIONS  AND  FORMULAS 


PAGE 

Parallel  planes     .      .      .      .321 
Perpendicular  planes       .      .    338 

Plane 319 

determination  of     .        319,320 
Planes,   parallel    .      .      .      .321 

Plato 491,493 

Polar    distance    of    circle      .   429 

triangle 441 

Pole 427 

Polygon,  spherical      .      .      .   438 
Polyhedral    angle        .      .      .349 

convex 349 

edges  of 349 

face  angles  of   ....   349 

faces  of 349 

vertex    of 349 

Polyhedral  angles,  equal  350 
symmetrical  ....  350 
vertical  .  .  .  .  .  .350 

Polyhedron 360 

convex 360 

diagonal  of 360 

edges  of 360 

faces  of 360 

regular          391 

section    «)f 360 

vertices  of 360 

Polyhedrons 497 

classification  of       ...    360 

similar 396 

Postulate  of  solid  geometry  320 

Prism 361 

altitude  of 361 

bases  of 361 

circumscribed    about    cyl- 
inder     405 

inscribed  in  cylinder  .  .  405 
lateral  area  of  .  .  .  .361 
lateral  edges  of  .  .  .361 
.lateral  faces  of  ...  361 


PAGE 

Prism,  oblique 362 

quadrangular  ....  362 

regular 362 

right 362 

right  section  of  .  .  .361 

triangular 362 

truncated 362 

Prismatoid 389 

bases  of 389 

formula  for  volume  of  .  390 

Prismoid 389 

Projection  of  line  on  plane  338 
of  point  on  plane  .  .  .  338 

Pyramid 377 

altitude  of  .  .  .  .  .  .  377 

axis  of 378 

base  of 377 

circumscribed  about  cone  414 
frustum  of  ....  378 
inscribed  in  cone  .  .  .414 
lateral  area  of  .  .  .  .  377 
lateral  edges  of  .  .  .377 
lateral  faces  of  ...  377 
quadrangular  ....  377 

regular 377 

regular,  slant  height  of  .  378 

spherical 461 

triangular  .  .  .  .  .  377 

truncated 378 

vertex  of 377 

Pythagoras   .   491,  492,  495,  497 

Radius  of  sphere  .  .  .  425 
Reciprocity,  principle  of  .  486 
Rectilinear  figures  .  .  .  494 
Rhetorical  methods  .  .  .491 
Right  section  of  cylinder  .  405 

of  prism 361 

Romans  ....  494,496 
Round  Bodies,  The  Three  498 


INDEX  OF  DEFINITIONS  AND  FORMULAS 


523 


PAGE 

Section  of  Polyhedron     .      .   360 

of  sphere 461 

Segment  of  sphere     .      .      .   462 

Similar   cones   of   revolution  413 

cylinders  of  revolution      .   404 

Sectors  of  circles        .      .      .275 

Segments  of  circles     .      .      .   275 

Slant  height  of  cone       .      .413 

of  frustum  of  cone       .      .414 

of  frustum  of  pyramid     .    378 

of  regular  pyramid      .      .377 

Solid  geometry      .      . '    .      .319 

Solids,  equivalent       .      .      .   363 

Spaces,  curved       ....   488 

Sphere 425 

axis  of 427 

center  of 425 

circumscribed  about  poly- 
hedron        432 

diameter    of       ....   425 
formula   for   area   of   sur- 
face of 455 

formula  for  volume  of,  .  463 
great  circle  of  ....  427 
inscribed  in  polyhedron  .  432 

poles  of 427 

radius  of 426 

small  circle   of        ...   427 
Spherical  angle     ....    433 

degrees 452 

excess      ....        444, 460 
Spherical   polygon      .      .      .   438 

angles  of 438 

sides  of 438 

vertices  of 438 

Spherical  pyramid      .      .      .   461 

base  of 461 

vertex  of 461 

Spherical  sector    .      .      .      .461 
base  of 461 


PAGE 

Spherical  sector,  formula  for 

volume  of       ....   464 

Spherical  segment       .      .      .   462 

altitude  of 462 

bases  of 462 

formulas  for  volumes  of    .   465 
of  one  base    ....       462 

Spherical  triangle  .  .  .  438 
bi-rectangular  ....  443 
formula  for  area  of,  in 

square  units  of  area      .   460 
tri-rectangular  ....   443 

Spherical    triangles,    supple- 
mental        442 

symmetrical       ....   444 

Spherical  wedge     .      .      .      .461 

Spherid 452 

Surface,  conical  .  .  .  .413 
cylindrical 403 


Tangent,  line  to  sphere 

plane  to  cone    . 

plane  to  cylinder    . 

plane  to  sphere 

spheres    .... 
Tetrahedral   angle 
Tetrahedron      . 
Thales   .      .      . 
Triangle,    polar 

spherical 
Trihedral  angle 

bi-rectangular 

isosceles 

rectangular 

tri-rectangular 


491, 


.  425 

.  413 

.  404 

.  426 

.  426 

.  349 

.  360 

495,  496 

.  441 

,  .  438 

.  349 

.  350 

.  350 

.  350 

.  350 


Ungula 461 

Unit  of  volume    ....   363 
Units  of  spherical  surface     .  452 


524 


INDEX  OF  DEFINITIONS  AND  FORMULAS. 


Vertex  of  cone 
of  polyhedral  angle     . 
of  pyramid 

PAGE 

.   412 
.   349 
.   377 

Wedge,  spherical 
Xenodorus 

PAGE 

.      .      .   461 
.      .      .  496 

of  spherical  pyramid  . 
Vertices  of  polyhedron 

.   461 
.   360 

.      .      .  452 

of  spherical  polygon 

.   438 

altitude  of   . 

452 

bases  of 

452 

Volume  of  solid     . 

.   363 

of  one  base  . 

.      .      .  452 

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